TS Inter 1st Year Chemistry Study Material Chapter 5 Stoichiometry

Telangana TSBIE TS Inter 1st Year Chemistry Study Material 5th Lesson Stoichiometry Textbook Questions and Answers.

TS Inter 1st Year Chemistry Study Material 5th Lesson Stoichiometry

Very Short Answer Type Questions

Question 1.
How many number of moles of glucose are present in 540 gms of glucose? [IPE ’14]
Answer:

Question 2.
Calculate the weight of 0.1 mole of sodium carbonate. [AP ’16]
Answer:
No. of moles of sodium carbonate

Question 3.
How many molecules of glucose are present in 5.23 g of glucose (Molecular weight of glucose 180 u).
Answer:

No. of molecules = No. of moles × Avogadro’s number
= \(\frac{5.23}{180}\) × 6.02 × 10²³ = 1.75 × 10²² molecules

Question 4.
Calculate the number of molecules present in 1.12 × 10^-7 c.c. of a gas at STP (c.c. – cubic centimeters = cm³).
Answer:
22400 cm³ contain 6.02 × 10²³ molecules
1.12 × 10^-7 cm³ contain ?

3 × 10¹² molecules.

Question 5.
The empirical formula of a compound is CH₂O. Its molecular weight is 90. Calculate the molecular formula of the compound. [AP ’16, Mar. ’13]
Answer:
Molecular formula = empirical formula × n

Empirical formula weight of CH₂O
= 12 + 2 + 16 = 30
n = \(\frac{90}{30}\) =3
Molecular formula = (CH₂O)₃ = C₃H₆O₃

Question 6.
Balance the following equation by the oxidation number method.
Cr_(s) + Pb(NO₃)_2(aq) → Cr(NO₃)_3(aq) + Pb_(S)
Answer:

Question 7.
What volume of H₂ at STP is required to reduce 0.795 g of CuO to give Cu and H₂O.
Answer:
H₂ reduces CuO according to the reaction
CuO + H₂ → Cu + H₂O
Moles of CuO = \(\frac{0.795}{79.5}\) = 0.01
Since 1 mol. of CuO can be reduced by 1 mol. of H₂
0.01 mol of CuO is reduced by 0.01 mol. of H₂
Volume of H₂ = 0.01 × 22.4 = 0.2242 lits.

Question 8.
Calculate the volume of 02 at STP required to completely burn 100 ml. of acetylene.
Answer:
C₂H₂ + \(\frac{5}{2}\) O₂ → 2 CO₂ +H₂O
To burn 22400 ml. of C₂H₂ the volume of O₂ required is 22400 × \(\frac{5}{2}\)
For burning 100 ml. of C₂H₂ the volume of O₂ requires
100 × 22400 × \(\frac{5}{2}\) × \(\frac{1}{22400}\) = 250 ml.

Question 9.
Now a days it is thought that oxidation is simply decrease in electron density and reduction is increase in electron density. How would you justify this?
Answer:
Oxidation involves loss of electrons whereas reduction involves gain of electrons. Thus oxidation is decrease in electron density whereas reduction is increase in electron density.

Question 10.
What is a redox concept? Give an example.
Answer:
Oxidation is the increase in oxidation number of the given species while the reduction is decrease in the oxidation number of the given species in a reaction. A chemical reaction in which both oxidation and reduction takes place simultaneously is called redox reaction.
e.g.: Na + \(\frac{1}{2}\)Cl₂ → NaCl

In the above reaction oxidation number of sodium increases from 0 to +1 while the oxidation number of chlorine decreases from 0 to -1. So sodium is oxidised and chlorine is reduced.

Question 11.
Calculate the mass percent of the different elements present in sodium sulphate (Na₂SO₄).
Answer:
Molecular weight of Na₂SO₄ = 142
142 gm of Na₂SO₄ contain 46 gms of sodium
∴ 100 gm of Na₂SO₄ contain \(\frac{100\times46}{142}\) = 32.38%
142 gm of Na₂SO₄ contain 32 gm of sulphur
∴ 100 gm of Na₂SO₄ contain \(\frac{100\times32}{142}\) = 22.54%
142 gm of Na₂SO₄ contain 64 gm of oxygen
∴ 100 gm of Na₂SO₄ contain \(\frac{100\times64}{142}\) = 45.08%

Question 12.
What do you mean by significant figures?
Answer:
Significant figures are meaningful digits which are known with certainity.
Eg: If we write 11.2 ml., the 11 is certain and 2 is uncertain and the uncertainity may be ±1 in the last digit. So the significant figure is 2.

Question 13.
If the speed of light is 3.0 × 10⁸ ms^-1. Calculate the distance covered hy light in 2.00 ns.
Answer:
Distance = speed × time
= 3 × 10⁸ × 2 × 10^-9 = 0.6 meter
So the distance covered by light in 2 ns = 0.6 meter.

Short Answer Questions

Question 1.
The approximate production of sodium carbonate per month is 424 × 10⁶ g. While that of methyl alcohol is 320 × 10⁶ gm. Which is produced more in terms of moles?
Answer:
Moles of sodium carbonate produced per month = \(\frac{424\times10^6}{106}\) = 4 × 10⁶
Moles of methyl alcohol produced per month = \(\frac{320\times10^6}{32}\) = 10⁷
So methyl.alcohol produced in terms of moles is more.

Question 2.
How much minimum volume of CO at STP is needed to react completely with 0.112 L of O₂ at 1.5 atm. pressure and 127°C to give CO₂.
Answer:
Reaction between CO and O₂
2 CO + O₂ → 2CO₂
Moles of O₂ = \(\frac{PV}{RT}=\frac{1.5\times0.112}{0.0821\times400}\) = 5.11 × 10^-3
According to the reaction for every one mole O₂ two moles of CO reacts.
∴ The minimum volume of CO required at
STP = 5.11 × 10^-3 × 2 = 10.22 × 10^-3
⇒ 10.22 × 10^-3 × 22400 = 229.32 ml.

Question 3.
Chemical analysis of a carbon compound gave following percentage composition by weight of the element present, carbon =10.06%, hydrogen = 0.84%, chlorine = 89.10%. Calculate the empirical formula of the compound.
Answer:

∴ Empirical formula of the compound = CHCl₃

Question 4.
A carbon compound on analysis gave the following percentage composition, carbon 14.5%, hydrogen 1.8%, chlorine 64.46%, oxygen 19.24%. Calculate the empirical formula of the compound.
Answer:

∴ Empirical formula of the compound = C₂H₃Cl₃O₂

Question 5.
Calculate the empirical formula of a compound having percentage composition:
Potassium (IQ = 26.57; Chromium (Cr) = 35.36, Oxygen (O) = 38.07.
(Given the Atomic weights of K, Cr and O are 39, 52 and 16 respectively)
Answer:

∴ Empirical formula of the compound = K₂Cr₂O₇

Question 6.
A carbon compound contains 12.8 % Carbon, 2.1 % Hydrogen, 85.1 % Bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula. [AP Mar. ’17]
Answer:

Empirical formula of the compound = C₁H₂Br₁
Empirical formula weight = 1 × C + 2 × H + 1 × Br = 1 × 12 + 2 × 1 + 1 × 80 = 94
Molecular weight = 187.9

Molecular formula = Empirical formula × n = C₁H₂Br₁ × 2 = C₁H₄Br₂.

Question 7.
0.188 g of an organic compound having an empirical formula CH₂Br displaced 24.2 cc of air at 14°C and 752 mm pressure. Calculate the molecular formula of the compound. (Aqueous tension at 14°C is 12 mm)
Answer:
Pressure of dry gas = Pressure of gas – aqueous tension = 752-12 = 740 mm
According to ideal gas equation PV = \(\frac{W}{M}\) RJ.

Substiuting these values in ideas gas equation
Molecular weight

Empirical formula of the compound = CH₂Br
Empirical formula wt. of the compound = 12 + 2 + 80 = 94

Question 8.
Calculate the amount of 90% H2S04 required for the preparation of 420 kg HCl.
2 NaCl + H₂SO₄ → Na₂SO₄ + 2HCl
Answer:
No. of moles of HCl to be prepared = \(\frac{420\times10^3}{36.5}\) = 11.5 × 10³

According to the reaction for every two moles of HCl one mole of H₂SO₄ is required. Therefore the no. of moles of H₂SO₄ required is

\(\frac{11.5}{2}\) × 10³ = 5.75 × 10³
Wt. of H₂SO₄ = 5.75 × 10³ × 98 = 563.5 kg
Since the given H₂SO₄ contain only 90%.
The weight to be taken is \(\frac{563.5\times100}{90}\) = 627 kg.

Question 9.
An astronaut receives the energy required in his body by the combustion of 34g of sucrose per hour. How much oxygen he has to carry along with him for his energy requirement in a day?
Answer:
Wt. of sucrose required per day = 34 × 24
= 816 gm
Moles of sucrose = \(\frac{W}{M.Wt.}=\frac{816}{342}\) = = 2.385
Sucrose react with oxygen as follows.
C₁₂H₂₂O₁₁ + 12O₂ → 12 CO₂ + 11 H₂O
According to the above reaction
1 mole sucrose requires – 12 moles of O₂
2.385 moles requires = \(\frac{2.385\times12}{1}\) = 28.63
Wt. of oxygen = No. of moles × Mol. Wt.
= 28.63 × 32 = 916.2 gm.

Question 10.
What volume of CO₂ is obtained at STP by heating 4 g of CaCO₃?
Answer:
Calcium carbonate decomposes on heating.

According the reaction
1 mole of CaCO₃ on heating gives 1 mole of CO₂
Mol. wt. of CaCO₃ = 100
∴ 100 gm of CaCO₃ on heating gives 22.4 lit.
4 gm of CaCO₃ on heating gives ?
\(\frac{4\times22.4}{100}\) = 0.896 lit.

Question 11.
When 50 gm of a sample of sulphur was burnt in air 4% of the sample was left over. Calculate the volume of air required at STP containing 21% oxygen by volume.
Answer:
Amount of Sulphur taken = 50 gm
Wt. of sulphur left = 4% = 2 gm
Wt. of sulphur reacted = 50 – 2 = 48 gm
Sulphur burns in air according to the reaction
S + O₂ → SO₂
Moles of Sulphur = \(\frac{48}{32}\) = 1.5
Moles of Oxygen required = 1.5
Volume of oxygen at STP = 22.4 × 1.5 = 33.6 lit.
Volume of air = \(\frac{33.6\times100}{21}\) =160 lit.
(∴ air is 21% O₂)

Question 12.
Calculate the volume of oxygen gas required at STP conditions for the complete combustion of 10 cc of methane gas at 20°C and 770 mm pressure.
Answer:
Methane burns according to the reaction
CH₄ + 2O₂ → CO₂ + 2H₂O

Moles of O₂ = 4 × 10^-4 × 2 = 8 × 10^-4
Vol. of O₂ at STP = 8 × 10^-4 × 22400
= 18.88 cc.

Question 13.
Calculate the volume of H₂ liberated at 27°C and 760 mm of Hg pressure by action by 0.6g magnesium with excess of dil HCl.
Answer:
Magnesium reacts with dilute hydrochloric acid as
Mg + 2 HCl → MgCl₂ + H₂
No. of moles of Mg = \(\frac{0.6}{24}\) = 0.025
No. of moles of H₂ = 0.025
(∴ 1 mole Mg liberates)
Ideal gas equation PV = nRT
P = 760 mm = 1 atm T = 27 + 273 = 300 K
V = ? n = 0.025
R = 0.0821
Substituting these values in ideal gas equation.

Question 14.
Explain the role of redox reactions in titrimetre processes and galvanic cells.
Answer:
Role of redox reactions in titrimetric quantitative analysis:
Titrimetric analysis involves two substances. They are (1) a solution of known concentration or a standard solution and (2) a solution of unknown concentration. The first solution is also known as Titrant. The second solution is also known as Titrand. The process of adding a standard solution to the titrand till the reaction is just complete is called titration. The point at which the titrand just completely reacts with the standard solution is called “equivalence point” or “end point.”

In redox reactions the completion of the titration is detected by a suitable method like (a) observing a physical change.
Ex : The light pink colour of KMnO₄ titrations.

(b) by using a reagent known as indicator which gives a clear visual change in its colour.
Ex (1) In Cr₂O7^-2 (dichromate) titrations, diphenyl amine is used as a reagent and at the end point it produces intense blue colour due to its oxidation by Cr₂O7^-2.
Ex (2) In the titration of Cu+2 with F (Iodometry)
2Cu⁺²_(aq) + 4I^–_(aq) → Cu₂I_2(S) + I_2(aq)

The I₂ formed in the redox reaction gives a deep blue colour with starch solution, added to the flask.

In this way redox reactions are taken as the basis for titrimetric analysis with MnO^–₄, Cr₂O^-2₇ etc. as oxidising agents and S₂O^-2₃ etc. as reducing agents.

Role of Redox reactions in galvanic cells :
When a zinc rod is kept in copper sulphate solution then the following redox reaction takes place.

In this redox reaction the transfer of electrons from Zn_(s) to Cu⁺²_(aq) takes place directly. The same transfer of electrons can also be done indirectly in a galvanic cell (Daniel cell).

Cells in which chemical energy is converted into electrical energy are called galvanic cells. Daniel cell is a best example for a galvanic cell. The Daniel cell consists of two beakers containing zinc rod dipped in ZnSO_4(aq) solution in one beaker and a copper rod dipped in CuSO_4(aq) solution in a second beaker. The two beakers are connected by an inverted U – tube, known as salt bridge. The two rods are connected by means of wires to the terminals of an ammeter. Redox reaction takes place in each of the beakers. Each beaker contains both oxidised and reduced forms of the respective species. The two types of species present together in each beaker is called a redox couple. Each beaker contains a redox couple. The oxidised and reduced forms are separated by a vertical line or a slash.

In the above arrangement the two redox couples are represented by Zn⁺²/Zn and Cu⁺²/Cu. As the metal is in two different oxidation states at the interface (say Zn/ Zn⁺²), some potential is developed, which is called electrode potential’. These electrode potentials are very useful in metallurgy, electroplating etc.

In this way redox reactions play an important role in galvanic cells.

Question 15.
Define and explain molar mass.
Answer:
Molar mass :
The mass of one mole of any substance in grams is called its molar mass.
Ex : Molar mass of sulphuric acid = 98 g.
Molar mass of hydrogen
= one gram for a gm atomic mass.
= two grams for a gm molecular mass.

Thus molar masses are atomic weights, molecular weights, formula weights etc. expressed in grams.

Gram atomic weight is atomic weight expressed in grams. Gram molecular weight is molecular weight expressed in grams.

Gram atom :
One gram atomic weight of a substance is known as gram atom.

Gram molecule:
One gram molecular weightj of a substance is known as gram molecule.

Mole :
It is the mass of a substance which contains Avogadro number of structural units.

1 mole = 1 gram molecule
= 1 gram molecular weight
= Mass of 6.023 × 10²³ molecules in grams.
1 mole = 1 gram atom
= 1 gram atomic weight
= Mass of 6.023 × 10²³ atoms in grams.

Question 16.
What are disproportionate reactions? Give example. [TS ’16, ’15; Mar. ’10]
Answer:
Chemical reactions in which the same element undergoes both oxidation and reduction simultaneously are known as disproportionation reactions.

Question 17.
What is comproportionation reactions? Give example.
Answer:
The reverse of disproportionation is comproportionation. In these reactions, two species with the same element in two different oxidation states form a simple product in which the element is in an intermediate oxidation state.

Question 18.
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Answer:

The ratio of Fe and 0 atoms = 0.67 : 1
Multiply with 3 to make integer = 2:3
Empirical formula of oxide of iron = Fe₂O₃

Question 19.
Calculate the mass of sodium acetate (CH₃COONa) required to make 500 ml. of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^-1.
Answer:
Molar mass of CH₃COO Na – 82.0245 g mol^-1
1 mol = 82.02459
0.375 mol = ?
0.375 × 82.0245 gm
1000 ml. contain 0.375 × 82.0245 g
500 ml. contain ?
= \(\frac{500}{1000}\) × 0.375 × 82.0245
= 15.375 gm.
∴ The mass of CH₃COONa present in 500 ml
= 15.375 gm.

Question 20.
What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L^-1 if 20 g are dissolved in enough water to make a final volume upto 2L?
Answer:
Molarity = mole per litre

Question 21.
How many significant figures are present in the following?
i) 0.0025, ii) 208, iii) 5005, iv) 126,000 v) 500.0, vi) 2.0034
Answer:
(i) 0.0025
No. of significant figures = 2

(ii) 208
No. of significant figures = 3

(iii) 5005
No. of significant figures = 4

(iv) 1,26,000
No. of significant figures = 6

(v) 500.0
No. of significant figures = 4

(vi) 2.0034
No. of significant figures = 5

Question 22.
Round up the following upto three significant figures:
i) 34.216, ii) 10.4107, iii) 0.04597, iv) 2808
Answer:
(i) 34.216 = 34.2
(ii) 10.4107 = 10.4
(iii) 0.04597 = 0.046
(iv) 2808 = 2.81 × 10³

Question 23.
Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one). Use the data given in the following table to calculate the molar mass of naturally occuriiig argon isotopes:

Isotope	Isotopic molar mass	Abundance
36Ar	35.96755 g mol-1	0.337%
38Ar	37.96272 g mol-1	0.063%
40Ar	39.9624 g mol-1	99.600 %

Answer:
(a) Mole fraction of ethanol = 0.04

Moles of ethanol n₁ = 0.04
No. of moles of water = 1 – 0.04 = 0.996
Wt. of water = 0.996 × 18 gm
Vol. of water = 0.996 × 18 ml.
Molarity of ethanol

Question 24.
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g Calculate 0) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
Answer:
No. of moles of CO₂ = \(\frac{3.38}{44}\) = 0.07682
No. of moles of H₂O = \(\frac{0.69}{18}\) = 0.03833
Ratio of the moles of CO₂ = H₂O
= 0.07682 : 0.03833 = 2 : 1
∴ The ratio of carbon and hydrogen atoms is 1 : 1 (because 1 CO₂ = H₂O)
Empirical formula = CH
10.0 L at STP weigh 11.6 g
22.4 L at STP weigh ?
\(\frac{22.4\times11.6}{10}\) = 26
∴ Molecular wt. of compound = 26
Empirical formula weight = 13
= \(\frac{26}{13}\) =2
Molecular formula = (EF) × 2 = C₂H₂

Question 25.
Calcium Carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction,
CaCO₃(s) + 2 HCl_(aq) → CaCl₂(aq) + CO₂ (g) + H₂O
What mass of CaCOs is required to react completely with 25 ml of 0.75 M HCl?
Answer:
CaCO₃ (s) + 2 HCl (aq) → CaCl₂ (aq) + CO₂ (g) + H₂O (I)
Moles of HCl = \(\frac{25\times0.75}{1000}\) = 0.01875
With 2 mol. of HC/ the mole of CaCO₃ react is 1 mol. with 0.01875 mol. of HCl the mole of CaCO₃ that react is
\(\frac{0.01875\times1}{2}\) = 0.009375
Wt. of CaCO₃ = 0.009375 × 100 = 0.9375 gm.

Question 26.
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction
HCl (aq) + MnO₂(s) → 2H₂O (l) + MnCl₂ (aq) + Cl₂(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Answer:
Moles of MnO₂ = \(\frac{5}{87}\) = 0.0574
The reaction between MnO₂ and HCl given is
4 HCl (aq) + MnO₂(s) → 2H₂O(l) + MnCl₂(aq) + Cl₂(g)

As per the above reaction for 1 mol. of MnO₂ 1 mol. of Cl₂ is produced by the reaction with 4 mol. of HCl.

∴ 1 mol. of MnO₂ react with 4 × 36.5 gmm HCl.
0.0574 mol. of MnO₂ react with? HCl

Question 27.
To 50 ml. of 0.1 N Na₂CO₃ solution 150 ml. of H₂O is added. Then calculate the normality of resultant solution.
Answer:
V₁ = 50 V₂ = 50 + 150 = 200
N₁ = 0.1 N₂ = ?
V₁N₁ = V₂N₂

Question 28.
Calculate the volume of 0.1 NH₂SO₄ required to neutralise 200 ml. of 0.2 N NaOH solution.
It is an acid base neutralisation reaction. Hence, at the neutralisation point. Number of equivalents of acid = Number of equivalents of base.
Answer:
Vol. of H₂SO₄ V₁ = ?
Volume of NaOH V₂ = 200 ml.
Normality of H₂SO₄ N₁ = 0.1
Normality of NaOH N₂ = 0.2 N
V₁N₁ = V₂N₂

Question 29.
Calculate normality of H₂SO₄ solutions if 50 ml of it completely neutralise 250 ml. of 0.1 N Ba(OH)₂ solutions.
Answer:
Vol. of H₂SO₄, V₁ = 50 ml.
Volume of Ba(OH)₂, V₂ = 250 ml.
Normality of H₂SO₄, N₁ = ?
Normality of Ba(OH)₂, N₂ = 0.1
V₁N₁ = V₂N₂

Question 30.
Calculate the volume of 0.1M KMnO₄ required to react with 100 ml. of 0.1 M H₂C₂O₄. 2H₂O solution in the presence of H₂SO₄.
Answer:
Potassium permanganate react with oxalic acid according to the reaction
2 KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → K₂SO₄ + 2 MnSO₄ + 8 H₂O + 10 CO₂
Vol. of KMnO₄, V₁ = ?
Volume of H₂C₂O₄, V₂ = 100 ml.
Molarity of KMnO₄ = 0.1 M
Molarity of H₂C₂O₄, M₂ = 0.1
No. of moles of KMnO₄ n₁ = 2
No. of moles of H₂C₂O₄, n₂ = 5

Question 31.
Assign oxidation number to the underlined elements in each of the following species.

Answer:
a) +5
b) +6
c) +5
d)+6
e) -1
f) +3
g) +6
h) +6

Question 32.
What are the oxidation number to the underlined elements in each of the following and how do you rationalise your results?
a) KI₃
b) H₂S₄O₆
c) Fe₃O₄
Answer:
a) KI₃ → K⁺ + I^–₃

I^–₃ ion is formed by combining I^-1 with I₂.

The average oxidation number is \(\frac{1}{3}\) but I^– is in -1 oxidation state while I₂ is zero oxidation state.

b) H₂S₄O₆ has the following structure

The oxidation states of S₂ and S₃ are zero but the oxidation states of S₁ and S₄ are +5.
The average oxidation state is \(\frac{10}{4}\) = 2.5

c) Fe₃O₄ contain FeO and Fe₂O₃
In FeO oxidation state of Fe in FeO = +2
In Fe₂O₃ oxidation state of Fe in Fe₂O₃ = +3
So average oxidation of Fe

Question 33.
Justify that the following reactions are redox reactions.
a) CuO(s) + H₂(g) → Cu(s) + H₂O (g)
b) Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
c) 4BCl3(g) + 3UA1H4(S) → 2B2H6 (g) + 3 LiCl (s) + 3 AIC13(S)
d) 2K(s) + F₂(g) → 2K⁺F^–(s)
e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
Answer:
a) CuO(s) + H₂(g) → Cu(s) + H₂O (g)
In this reaction the oxidation number of Cu decreased from +2 to 0 and the oxidation state of H2 is increased to +1.
So it is a redox reaction.

b) Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
In this reaction the oxidation number of Fe ion Fe₂O₃ is decreased to zero in Fe from +3 and the oxidation number of carbon in CO is increased from +2 to +4 ion CO₂. So it is a redox reaction.

c) 4BCl₃(g) + 3ULiAlH₄(S) → 2B₂H₆(g) + 3 LiCl(s) + 3 AlCl₃(S)
In LiAlH₄, hydrogen is present as H^– ion with more negative charge on H. But ion B₂ H₆ ; also the H atom will have some negative charge as the electronegativity of H is 2.1 ; while that of boron is 2.0.

According to the modern concept decrease in electron density is reduction nnd increase in electron density is oxidation.

Here the electron density decreases at hydrogen and increases at boron because (he bond with more electronegative atom i (B – Cl) changes to less electronegative ; atom (B – H). So it is also redox reaction.

d) 2K(s) + F₂(g) → 2K⁺F^–(s)
In the formation of K⁺F^–, K loses electron (oxidation) and F gains electron (reduction) so it is redox reaction.

e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
The oxidation of N increases from -3 to +2 in the conversation of NH₃ to NO. It is oxidation.

The oxidation number of 02 changes from zero to -2.

It is reduction. So it is redox reaction.

Question 34.
Fluorine reacts with ice and results in the change.
H₂O (S) + F₂(g) → HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
Answer:
The electron density at O – atom decreases when the O – H bond changes to O – F since electronegativity of F is more than H. The decrease in electron density is reduction.

The electron density in F₂ is zero, but in HOF the electron density in F increases. The increase in electron density is reduction. So the above reaction is redox reaction.

Question 35.
Calculate the oxidation number of sulphur, chromium and nitrogen ion H₂SO₅, Cr₂O^2-₇ and NO^–₃. Suggest structure of those compounds.
Answer:
H₂SO₅
Oxidation number of H = +1
Oxidation number of O = -2
Oxidation number of S = x
(2 × 1) + x + 3(-2) + 2(—1) = 0
2 + x – 6 – 2 = 0
x – 6 = 0
x = + 6

i) The oxidation number of sulphur is exceeding its group number which cannot exist. So it should contain peroxy bond.

ii) Cr₂C^2-₇
Oxidation state of chromium = x
Oxidation state of oxygen = – 2
(2x) + (-2 × 7) = -2
2x = + 12
∴ x = \(\frac{+12}{2}\) = +6
The oxidation state Cr = +6

(iii) NO^–₃
Oxidation state of N = x
Oxidation state of 0 = -2
x + (-2 × 3) = -1
x = +5

Question 36.
Write the formulae for the following compounds.
a) Mercury (II) chloride
b) Nickel (II) sulphate
c) Tin (IV) oxide
d) Thallium (I) sulphate
e) Iron (III) sulphate
f) Chromium (III) oxide.
Answer:
a) HgCl₂
b) NiSO₄
c) Sn₂O₄
d)Tl₂SO₄
e) Fe₂(SO₄)₃
f) Cr₂O₃

Question 37.
Suggest a list of the substances where carbon exhibit oxidation states from – 4 to 4 and nitrogen from -3 to +5.
Answer:
List of carbon compounds that exhibit oxidation states from -4 to +4
The underlined carbon in the following compounds have the oxidation*state mentioned.

Question 38.
While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing |gents in their reactions, ozone and nitric acid act only as oxidants. Why?
Answer:
In sulphur dioxide sulphur is in +4 oxidation state. It can increase its oxidation number upto +6 while acting as reducing agent and can decrease its oxidation number upto either 0 or -2 while acting as oxidising agent.

Similarly in hydrogen peroxide oxidation number of oxygen is -1. It can increase its oxidation number upto zero and can decrease its oxidation number to -2.

Therefore SO₂ and H₂O₂ can act as oxidising and reducing agents in their reactions.

In ozone the oxidation number of oxygen is zero. It can only decrease its oxidation number but cannot increase its oxidation number. This is because it is only the most electronegative atom next to fluorine.

In nitric acid oxidation state of nitrogen is +5. It cannot increase its oxidation state because it is the maximum oxidation state of nitrate. It can only decrease its oxidation number.

Because of these reasons ozone and nitric acid can act only as oxidising agents.

Question 39.
Consider the reactions
a) 6CO₂(g) + 6H₂O (I) → C₆H₁₂O₆(aq) + 6O₂(g)
b) O₃(g) + H₂O₂ (I) → H₂O (I) + 2O₂ (g)
Why it is more appropriate to write these reactions as
a) 6CO₂(g) + 12H₂O (I) → C₆H₁₂O₆ + 6H₂O(I) + 6O₂(g)
b) O₃(g) + H₂OZ (I) → H₂O (I) + O₂ (g) + O₂(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Answer:
Plants absorb carbon dioxide from air, water from soil and convert them into carbohydrates in the presence of sunlight and Chlorophyll. This process is known as photosynthesis.

During photosynthesis plants liberate oxygen. The oxygen will be liberated from water but not from carbon dioxide. The following reaction cannot explain the liberation of oxygen from water because in this reaction from 6H₂O molecules only 3O₂ can be liberated.
6CO₂(g) + 6H₂O → C₆H₁₂O₂(aq) + 6O₂(g)

But the following reaction can explain the liberation 6O₂ molecules from water.
6CO₂(g) + 12H₂O (I) → C₆H₁₂O₆ (aq) + 6H₂O(l) + 6O₂ (g)

The path of the reaction can be traced by taking labile 0 in H₂O. The liberated oxygen contain the total labile ¹⁸O which indicates the oxygen is liberated from water.
6CO₂(g) + 12H₂¹⁸O(l) → C₆H₁₂O₆ (aq) + 6H₂O(l) + 6 ¹⁸O₂ (g)

b) The reaction between O₃ and H₂O₂ can be written as follows :

So it is appropriate to the equation as above instead of
O₃ + H₂O₂ → H₂O + 2O₂

In the reaction O₃ + H₂O₂ → H₂O + O₂ + O₂

One of the O₂ liberated from O₃ and the another from H₂O₂. This can be traced by using ¹⁸O isotope in H₂O₂.

Question 40.
The compound AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why?
Answer:
AgF₂ is unstable. So it dissociate into AgF and F. The fluorine liberated is a strong oxidising agent. So AgF₂ is strong oxidising agent. The Ag present in AgF₂ is in +2 oxidation state. This unstable Ag²⁺ will be reduced to stable Ag⁺ during this reaction.

Question 41.
Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
Answer:
1) In the reaction between HgCl₂ and SnCl₂, HgCl₂ act as oxidising agent and SnCl₂ act as reducing agent. If SnCl₂ is excess the product Hg is in its lower oxidation state. But if HgCl₂ is excess the product , is Hg2Cl₂

2) In the reaction between phosphorous and chlorine phosphorous is reducing agent and chlorine is oxidising agent. If chlorine is in small amount the product is PC/3 but in the presence of excess chlorine PCl₅ is the product.
P₄ + 6Cl₂ → 4 PCl₃
P₄ + 10Cl₂ → 4 PCl₄

3) When chlorine is passed into excess of liquid sulphate the product is sulphur monochloride S₂Cl₂. But if excess chlorine is passed until it is saturated, the product is SCl₂.

Question 42.
How do you count the following observations?
a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write balanced redox equation for the reaction.
b) When concentrated sulphuric acid is added to inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?
Answer:
a) Acidified permanganate oxidises organic compounds to carbon dioxide and water. But alkaline permanganate oxidises the organic compounds to aldehydes and acids.

So for the manufacture of benzoic acid from toluene alkaline permanganate is used instead of acidified permanganate.
2KMnO₄ + H₂O → 2MnO₂ + 2KOH + 3(O)
C₆H₅CH₃ + 3(O) → C₆H₅COOH + H₂O

2KMnO₄ + C₆H₅CH₃ → C₆H₅COOH + 2MnO₂ + 2KOH

b) Less volatile acids substitute more volatile acids from the salts. Concentrated sulphuric acid is less volatile and can substitute more volatile HCl and HBr from chlorides and bromides. But HBr is a reducing agent while HCl cannot act as reducing agent. So sulphuric acid can oxidise the colourless HBr to red vapour of bromine.
2 NaCl + H₂SO₄ → Na₂SO₄ + 2 HCl
2 KBr + H₂SO₄ → K₂SO₄ + 2 HBr
2 HBr + H₂SO₄ → 2H₂O + SO₂ + Br₂

Question 43.
Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:
a) 2 AgBr (s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr (aq) + C₆H₄O₂(aq)
b) HCHO (l) + 2 [Ag (NH₃)₂)⁺ (aq) + 3OH^–(aq) → 2 Ag(s) + HCOO^– (aq) + 4NH₃ (aq) + 2H₂O(l)
c) HCHO (l) + 2Cu²⁺ (aq) + 50H^– (aq) → Cu₂O (s) + HCOO^–(aq) + 3H₂O (l)
d) N₂H₄ (l) + 2H₂O₂ (l) → N₂(g) + 4H₂O (l)
e) Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → PbSO₄(s) + 2H₂O(l)
Answer:
A substance which undergoes oxidation acts as a good reducing agent while the one which undergoes reduction acts as a good oxidising agent.

a) Oxidising agent is AgBr and reducing agent is C₆H₆O₂.
b) Oxidising agent is ammonical silver nitrate (Tollen’s reagent) while reducing agent is HCHO.
c) Cu²⁺ undergoes reduction. So it is oxidising agent HCHO undergoes oxidation. So it is reducing agent.
d) Nitrogen in N₂H₄ undergoes oxidation. So it is reducing agent.
H₂O₂ undergoes reduction. So it is oxidising agent.
e) Pb undergoes oxidation. So it is reducing agent. PbO₂ undergoes reduction. So it is oxidising agent.

Question 44.
Consider the reactions
Why does the same reductant, thiosulphate react differently with iodine and bromine?
Answer:
Iodine is a weak oxidising agent while bromine is stronger oxidising reaction. So the oxidation of S₂O^2-₃ with iodine will take place until the oxidation state of sulphur +2 in S₂O^2-₃ changes to 2.5 in S₄O^2-₆ only. But bromine being stronger oxidising agent can oxidise the sulphur ion S₂O^2-₈ to its highest oxidation state +6 in S0^2-₄.

Question 45.
Justify giving reactions that among halo-gens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer:
Among halogens oxidation power decreases from fluorine to iodine due to decrease in electronegativities and electron gain enthalpies. This can be explained as follows.

Fluorine can displace Cl₂, Br₂ and I₂ from the corresponding halides.
2KCl + F₂ → 2KF + Cl₂
2KBr + F₂ → 2KF + Br₂
2KI + F₂ → 2KF + I₂

Chlorine can displace Br₂ and I₂ from bromides and iodides respectively but cannot displace fluorine from fluorides
2KBr + Cl₂ → 2KCl + Br₂
2KI + Cl₂ → 2KCl + I₂

Bromine can displace I₂ from iodide but cannot displace F2 from fluorides or C/2 from chlorides.
2KI + Br₂ → 2KBr + I₂

Iodine cannot displace any other halogen from their halides.

In the hydrogen halides the reduction power increases from HF to HI. This is because of the decrease in thermal stability of hydrogen halides with increase in bond length. Further the tendency to hold the electron decreases from HF to HI. So HF cannot be oxidised but HI can be easily oxidised. Hence HI is the best reductant.

Question 46.
Why does the following reaction occur?
XeO^4-₆ (aq) + 2 F^– (aq) + 6H⁺ (aq) → XeO₃(g) + F₂(g) + 3H₂0 (l)
What conclusion about the compound Na₄XeO₆ (of which XeO^4-₆ is a part) can be drawn from the reaction.
Answer:
The perxenate ion XeO^4-₆ ion is very strong oxidising agent than fluorine. So it can oxidise F^– ion to fluorine in acid medium. Hence the reaction occurs.
XeO^4-₆ (aq) + 2F^– (aq) + 6H⁺ (aq) → XeO₃(g) + F₂ (g) + 3H₂O(l)

Question 47.
Consider the reactions:
a) H₃PO₂ (aq) + 4AgNO₃ (aq) + 2H₂O (l) → H₃PO₄ (aq) + 4Ag(s) + 4HNO₃ (aq)
b) H₃PO₂ (aq) + 2CuSO₄ (aq) + 2H₂O (l) → H₃PO₄ (aq) + 2Cu(s) + H₂SO₄(aq)
c) C₆H₅CHO(l) + 2[Ag(NH₃)₂]⁺ (aq) + 3OH^–(aq) → C₆H₅COO^–(aq) + 2Ag(s) + 4NH₃(aq) + 2H₂O (l)
d) C₆H₅CHO (l) + 2 Cu²⁺ (aq) + 5OH^– (aq) → no change is observed.
What inference do you draw about the behaviour of Ag+ and Cu2+ from these reactions ?
Answer:
Ag⁺ and Cu²⁺ both can oxidise H₃PO₂ in acid medium but Ag⁺ oxidises H₃PO₂ to H₃PO₃. While Cu²⁺ is oxidising H₃PO₂ to H₃PO₄. Cu²⁺ is oxidising phosphorous H₃PO₂ from +1 to +5 oxidation state but Ag⁺ is oxidising +1 to +3. This indicates that Cu²⁺ is acting as strong oxidising agent than Ag⁺ in acid medium.

In alkaline medium Ag⁺ is oxidising benzaldehyde to benzoate but Cu²⁺ has no action. This indicates that in alkaline medium Ag⁺ is stronger oxidising agent than Cu²⁺.

Question 48.
Balance the following redox reactions by ion – electron method. [AP ’15; IPE ’14]
a) MnO^–₄ + I^– (aq) → MnO₂(s) + I₂(s) (In basic medium)
b) MnO^–₄ + SO₂ (g) → Mn²⁺ (aq) + HSO₄ (aq) (in acidic solution)
c) H₂O₂ (aq) + Fe³⁺ (aq) → Fe³⁺ (aq) + H₂O (l) (in acidic solution)
d) Cr₂O^2-₇ + SO₂ (g) → Cr³⁺ (aq) + SO₄^2- (aq) (in acidic solution) [Mar. ’18 AP]
Answer:
a) MnO^–₄ + I^– (aq) → MnO₂(s) + I₂(s) (In basic medium)

b) MnO^–₄ + SO₂ (g) → Mn²⁺ (aq) + HSO₄ (aq) (in acidic solution)

c) H₂O₂ (aq) + Fe³⁺ (aq) → Fe³⁺ (aq) + H₂O (l) (in acidic solution)


d) Cr₂O^2-₇ + SO₂ (g) → Cr³⁺ (aq) + SO₄^2- (aq) (in acidic solution) [Mar. ’18 AP]

Question 49.
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
a) P₄(s) + OH^– (aq) → PH₃(g) + H₂PO^–₂ (aq)
b) N₂H₄(l) + ClO^–₃ (aq) → NO (g) + Cl^–(g)
c) Cl₂O₇(g) H₂O₂(aq) → ClO^–₂(aq) + O₂(g) + H⁺
Answer:
a) P₄(s) + OH^– (aq) → PH₃(g) + H₂PO^–₂ (aq)
Ion electron method:

Note : Here P₄ acts both as oxidant and reductant.

Oxidation number method:

In order to balance the change in oxidation number H₂PO₂^– is to be multiplied by 3
P₄ + OH^– → PH³ + 3H₂PO^–₂

Since the reaction is taking place in basic medium, H₂O is to be added on the side which has lesser H atoms and OH” are to be added on the side which has lesser O atoms.
P₄ + 3H₂O + 3OH^– → PH₃ + 3H₂PO^–₂

b) N₂H₄(l) + ClO^–₃ (aq) → NO (g) + Cl^–(g)

Step – III: Equalise the increase and decrease in ON by multiplying N204 with 3 and C103 with 4.
3N₂O₄ + 4 ClO^–₃ → 6NO + 4 Cl^–

Step – IV: Balance the atoms except H and O. Here they are balanced.
Step – V : Balance O atoms by adding OH^– ions and H atoms by adding H20 on the sides deficient of O and H atoms respectively
3N₂O₄ + 4 ClO^–₃ → 6NO + 4 Cl^– + 12OH^–

c) Cl₂O₇(g) H₂O₂(aq) → ClO^–₂(aq) + O₂(g) + H⁺
Ion electron method:

Oxidation number methed:

Step – II : Equalise the increases / decrease in ON by multipling H₂O₂ with 4 since in each chlorine of Cl₂O₇ decrease in ON is 4. For 2 Cl atoms it is 8. In H₂O₂ increase in ON for each 0 is 1 and for two 0 atoms it is 2.
Cl₂O₇ + 4H₂O₂ → 2ClO^–₂ + 4H₂O + 2O₂

Step-III : Balance the O atoms by adding OH^– and H atoms by adding H20 to the sides deficient of O and H atoms respectively.
Cl₂O₇ + 4H₂O₂ + 2OH^– → 2ClO¹₂ + 4H₂O + 2O₂

Question 50.
What sorts of informations can you draw from the following reaction?
(CN)₂(g) + 2OH^– (aq) → CN^– (aq) + CNO^– (aq) + H₂O (l)
Answer:
In the above reaction the cyanogen gas undergoes disproportionation in basic medium. Here the oxidation state of CN radical decreases to -1 in CN- and increases to +1 in CNO” in basic medium.

Question 51.
The Mn³⁺ ion is unstable solution and undergoes disproportionation to give Mn²⁺, MnO₂ and H⁺ ion. Write balanced ionic equation for the reaction.
Answer:
Mn³⁺ + 2H₂O → MnO₂ + Mn²⁺ + 4H⁺

Question 52.
Consider the elements Cs, Ne, I and F.
a) Identify the element that exhibits only negative oxidation state.
b) Identify the element that exhibits only positive oxidation state.
c) Identify the element that exhibit both positive and negative oxidation states
d) Identify the element which neither exhibit the negative nor does the positive oxidation state.
Answer:
a) ‘F’ exhibit only negative oxidation state because it is the most electronegative atom.

b) Cs’ exhibit only positive oxidation state because it is the most electropositive element.

c) I can exhibit both positive and negative oxidation states. Ex : In ICl₃ the oxidation state of I is +3 and in Nal oxidation state of I is -1.

d) Ne being inert gas do not participate in reactions. So it will not exhibit neither the negative nor the positive oxidation states.

Question 53.
Chlorine is used to purify drinking water. Excess of Chlorine is harmful. The excess of Chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
Answer:
SO₂ + Cl₂ + 2H₂O → H₂SO₄ + 2HCl

Question 54.
Refer to the periodic table given in your book and now answer the following questions.
a) Select the possible non metals that can show disproportionation reaction
b) Select the metals that can show disproportionation
Answer:
a) Phosphorous, sulphur, chlorine, bromine, iodine
b) Copper, silver, gold

Question 55.
In Ostwal’s process for the manufacture of nitric acid the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen.
Answer:
The oxidation of ammonia to NO in Ostwalds process can take place as follows.
4NH₃ + 5O₂ → 4NO + 6H₂O + energy

68 gm of ammonia react with 160 gm of Oxygen. In this reaction oxygen is limiting reagent. Since to react with 10 g of ammonia the required amount of oxygen is \(\frac{10\times160}{68}\) = 23.53 gm of oxygen is required

But there is only 20.00 g of oxygen.
160 gm of O₂ can react with 68 gm of NH₃
∴ 20 gm of O., can react with \(\frac{20\times68}{160}\)
= 8.5 gm NH₃
For 68 gm of NH₃ the wt. of NO formed is 120
For 8.5 gm of NH₃ the wt. of NO formed is 15 gm.

Question 56.
i) Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn
ii) Calculate the molarity of sodium carbonate in a solution prepared by dissolving 5.3 g in enough water to form 250 ml of the solution. [Mar. ’13]
Answer:

Long Answer Questions

Question 1.
Write the balanced ionic equation which represents the oxidation of iodide (I^–) ion by per manganate ion in basic medium to give iodine (I) and manganese dioxide (MnO₁). [IPE ’14 AP Mar. ’19]
Answer:

Question 2.
Write the balanced ionic equation for the oxidation of sulphite ions to sulphate ions in acid medium by permanganate ion.
Answer:

Question 3.
Oxalic acid is oxidised by permanganate ion in acid medium of Mn²⁺ balance the reaction by ion-electron method. (Board Paper)
Answer:

Question 4.
Phosphorus when heated with NaOH solution gives Phosphine (PH₃) and H₂PO^–₂ Give balanced equation.
Answer:
a) P₄(s) + OH^– (aq) → PH₃(g) + H₂PO^–₂ (aq)
Ion electron method:

Note : Here P₄ acts both as oxidant and reductant.

Oxidation number method:

In order to balance the change in oxidation number H₂PO₂^– is to be multiplied by 3
P₄ + OH^– → PH³ + 3H₂PO^–₂

Since the reaction is taking place in basic medium, H₂O is to be added on the side which has lesser H atoms and OH” are to be added on the side which has lesser O atoms.
P₄ + 3H₂O + 3OH^– → PH₃ + 3H₂PO^–₂

b) N₂H₄(l) + ClO^–₃ (aq) → NO (g) + Cl^–(g)

Step – III: Equalise the increase and decrease in ON by multiplying N204 with 3 and C103 with 4.
3N₂O₄ + 4 ClO^–₃ → 6NO + 4 Cl^–

Step – IV: Balance the atoms except H and O. Here they are balanced.
Step – V : Balance O atoms by adding OH^– ions and H atoms by adding H20 on the sides deficient of O and H atoms respectively
3N₂O₄ + 4 ClO^–₃ → 6NO + 4 Cl^– + 12OH^–

c) Cl₂O₇(g) H₂O₂(aq) → ClO^–₂(aq) + O₂(g) + H⁺
Ion electron method:

Oxidation number methed:

Step – II : Equalise the increases / decrease in ON by multipling H₂O₂ with 4 since in each chlorine of Cl₂O₇ decrease in ON is 4. For 2 Cl atoms it is 8. In H₂O₂ increase in ON for each 0 is 1 and for two 0 atoms it is 2.
Cl₂O₇ + 4H₂O₂ → 2ClO^–₂ + 4H₂O + 2O₂

Step-III : Balance the O atoms by adding OH^– and H atoms by adding H20 to the sides deficient of O and H atoms respectively.
Cl₂O₇ + 4H₂O₂ + 2OH^– → 2ClO¹₂ + 4H₂O + 2O₂

Question 5.
Balance the following equation.

Answer:

Question 6.
Balance the following equation by the oxidation number method.
MnO^2-₄ + Cl₂ → MnO^2-₄ + Cl^–
Answer:
Step -1: The skeleton reaction
MnO^2-₄ + Cl₂ → MnO^2-₄ + Cl^–

Step – III: Equalise the increase / decrease in ON. Here they are equal.

Step – IV : Balance the other atoms except HandO
2MnO^2-₄ + Cl₂ → 2MnO^–₄ + 2Cl^–

Step – V : Balance H atoms and 0 atoms. Here they are balanced.
The balanced equation is
2MnO2^2-₄ + Cl₂ → 2MnO^–₄ + 2Cl^–

Question 7.
Explain the different types of redox reac-tions.
Answer:
A chemical reaction in which both oxidation and reduction reactions are involved is called an oxidation – reduction reaction or simply a redox reaction.

In this reaction Zn loses two electrons to form Zn⁺² and undergoes oxidation. Cu⁺² gains two electrons to form Cu and thus undergoes reduction.

Most of the chemical reactions are redox reactions. There are mainly four types of redox reactions. They are
(a) Chemical combination reactions
(b) Chemical decomposition reactions
(c) Chemical displacement reactions and
(d) Chemical disproportionation reactions

a) Chemical combination reactions:
Ex : Burning of coal in air.

In this reaction the oxidation numbers of carbon and oxygen are zero. In C02, the oxidation number of C is + 4 and that of oxygen is – 2. As the oxidation number of carbon increases from 0 to +4, we say that carbon undergoes oxidation. Similarly the oxidation number of oxygen decreases from 0 to – 2. Hence the oxygen undergoes reduction. Since this reaction involves both oxidation and reduction, we can infer that the above chemical combination reaction is a redox reaction.

b) Chemical decomposition reactions:
Ex. : Thermal decomposition of mercuric oxide.

In HgO, the oxidation number of Hg is +2 and that of oxygen is – 2. The oxidation numbers of free metallic mercury and elemental oxygen are zero. In this reaction Hg undergoes reduction from + 2 to 0 and oxygen undergoes oxidation from – 2 to 0. The decomposition of HgO involves both oxidation and reduction. Hence, we can infer that the above decomposition reaction is a redox reaction.

c) Chemical displacement reactions:
Ex. Zinc displaces copper from aqueous copper sulphate solution.

In this reaction, the oxidation numbers of elemental zinc and elemental copper are each zero, and the oxidation numbers of Cu and Zn in their aq. solutions are each +2. In this reaction the conversion of Zn into ZnSO₄ is oxidation and the conversion of CuSO₄ into Cu is reduction. Hence, we can infer that the above displacement reaction is a redox reaction.

d) Chemical disproportionation reactions:
Ex.: Chlorine is passed into cold and dilute solution of NaOH.

In this reaction the oxidation number of elemental chlorine is zero. The oxidation number of Cl in NaOCl is (+1) and in NaCl is (-1). In this reaction the same element chlorine has undergone both oxidation and reduction. Hence this is a redpx reaction.

Question 8.
State the law of definite proportions. Sug-gest one problem to understand the law by working out that problem.
Answer:
Law of definite proportions :
“A given chemical substance always contains the same elements combined in a fixed proportion by weight.”

Explanation :
SO₂ can be obtained by the following two methods.
i) When mercuric sulphide is heated in air, it gives mercury and sulphur dioxide.
HgS + O₂ → Hg + SO₂

ii) When lead sulphide is heated strongly in air, it gives lead oxide and sulphur dioxide.
2PbS + 3O₂ → 2PbO + 2SO₂

Samples of SO₂ obtained by the above two methods were analysed. In each of them, 100 g of SO₂ was found to contain 50 g of sulphur and 50 g of oxygen.

The above observations prove that the weight composition of sulphur dioxide is always constant.

Question 9.
How are the end points of titrations detected in the following reactions?
a) MnO^-2₄ oxidises Fe²⁺
b) Cr₂O^2-₇ oxidises Fe²⁺
c) Cu⁺² oxidises I^–
Answer:
a) In the oxidation of Fe²⁺ with MnO^–₄, the permanganate itself act as self indicator. MnO^–₄ has purple colour. The visible end point in this case is achieved after the last amount reductant (Fe²⁺) is oxidised and the first stable tinge of pink colour appears.

b) In the oxidation of Fe²⁺ with Cr₂O^2-₂ an indicator such as diphenyl amine is used. Just after the equivalence point the excess Cr₂O^2-₂ oxidises the diphenyl amine to intence blue colour by which the end point can be detected.

c) In the oxidation of I^– with Cu²⁺ the iodine formed will give intense blue colour with starch. This colour will be discharged with excess of hypo added after the equivalence point.

Question 10.
Calculate the amount of Carbondioxide that could be produced when
i) 1 mole of carbon is burnt in air
ii) 1 mole of carbon is burnt in 16g of dioxygen
iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Answer:

For burning 12g (1 mole) of carbon 32 gm of dioxygen is required. Since 16 g of dioxygen is present only 6 gm (half mole) of carbon burn producing half mole of CO₂.

Thus 22 g of CO₂ is formed.

(iii) Here also 22 g of CO₂ is formed since there is only 16 g of oxygen.

Question 11.
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation.
N₂(g) + H₂(g) → 2NH₃(g)
i) Calculate the mass of ammonia produced if 2.00 × 10³g dinitrogen reacts with 1.00 × 10³ g of dihydrogen.
ii) Will any of the two reactants remain unreacted?
iii) If yes, which one and what would be its mass?
Answer:
i) The balanced equation for the reaction between dihydrogen and dinitrogen is

When 28 g of N₂ react with 6 g of H₂ produce 34 g of NH₃

1 mole of N₂ can react with 3 moles of H₂
71.4 moles of N, can react = \(\frac{71.4\times3}{1}\)
= 214.2 moles of H₂

Here 1 mole of N₂ can produce 34 g of NH₃
71.4 mole of N₂ can produce
71.4 × 34 = 2427.6 gm.

iii) Here No. of moles of H₂ are more than required
The no. of moles of H₂ unreacted = 500 – 214.2 = 285.8
The amount of hydrogen left = 285.8 × 2 = 571.6 gms

Question 12.
Assign oxidation number to the underlined elements in each of the following species.

Answer:
a) +5
b) +6
c) +5
d) +6
e) -1
f) -5
g) +6
h) +6

Question 13.
What are the oxidation numbers of the underlined elements in each of the follow-ing and how do you rationalise your resuits?

Answer:
a) H₂S₄O₆ : H₂S₄O₆ has the following structure

The oxidation states of S₂ and S₃ are zero but the oxidation states of S₁ and S₄ are +5.

The average oxidation state is \(\frac{10}{4}\) = 2.5

b) Fe₃O₄ : Fe₃O₄ contain FeO and Fe₂O₃
In FeO oxidation state of Fe in FeO = +2
In Fe₂O₃ oxidation state of Fe in Fe₂O₃ = +3
So average oxidation of Fe = \(\frac{(+2)+2 \times(+3)}{3}=\frac{8}{3}=2.67\)

c) CH₃–CH₂-OH

When atoms of the same element combine their oxidation states are taken as zero. The carbon in CH₃ group is in -3 oxidation state.

The carbon in CH₂OH group is in zero oxidation state.

d) CH₃–COOH

The carbon in CH₃ is in -3 oxidation state while the carbon in COOH group is in +3 oxidation state.

Additional Questions & Answers

Question 1.
Calculate molecular mass of glucose (C₆H₁₂O₆) molecule.
Answer:
Molecular mass of glucose (C₆H₁₂O₆)
= 6(12.011 u) + 12.(1.008 u) + 6(16.00 u)
= (72.066 u) + (12.096 u) + (96.00 u)
= 180.162 u

Question 2.
A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?
Answer:
Step 1 : Conversion of mass per cent to grams:
Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen is present, 24.27g carbon is present and 71.65 g chlorine is present.

Step 2 : Convert into number moles of each element:
Divide the masses obtained above by respective atomic masses of various elements.

Step 3 : Divide the mole value obtained above by the smallest number:
Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl. In case the ratios are not whole numbers, then they may be con-verted into whole number by multiplying by the suitable coefficient.

Step 4:
These numbers indicate the rela+ tive number of atoms of the elements. Write empirical formula by mentioning the numbers after writing the symbols of respective elements :
CH₂Cl is, thus, empirical formula of the above compound.

Step 5: Writing molecular formul:
(a) Determine empirical formula mass. Add the atomic masses of various atoms present in the empirical formula.
For CH₂Cl, empirical formula mass is 12.1 + 2 x 1.008 + 35.453 = 49.48 g

(b) Divide molar mass by empirical formula mass

(c) Multiply empirical formula by n obtained above to get the molecular formula Empirical formula = CH₂Cl,
n = 2. Hence molecular formula is C₂H₄Cl₂.

Question 3.
Calculate the amount of water (g) produced by the combustion of 16 g of methane.
Answer:
The balanced equation for combustion of methane is :
CH₄(g) + 2O₂(g) → CO₂ (g) + 2H₂O (g)
(i) 16 g of CH₄ corresponds to one mole.
(ii) From the above equation, 1 mol of CH₄ (g) gives 2 mol of H₂O (g).
2 mol of water (H₂O) = 2 × (2 + 16)
= 2 × 18 = 36 g

Question 4.
How many moles of methane are required to produce 22 g CO₂ (g) after combustion?
Answer:
According to the chemical equation,
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)
44g CO₂ (g) is obtained from 16 g CH₄ (g).
[∵ 1 mol CO₂(g) is obtained from 1 mol of CH₄(g)].
mole of CO₂ (g)

Hence, 0.5 mol CO₂ (g) would be obtained from 0.5 mol CH₄ (g) or 0.5 mol of CH₄ (g) would be required to produce 22 g CO₂(g).

Question 5.
50.0 kg of N₂ (g) and 10.0 kg of H₂ (g) are mixed to produce NH₂ (g). Calculate the NH₂ (g) formed. Identify the limiting reagent in the production of NH₃ in this situation.
Answer:
A balanced equation for the above reaction is written as follows :
Calculation of moles:

According to the above equation, 1 mol N₂ (g) requires 3 mol H₂ (g), for the reaction. Hence, for 17.86 × 10² mol of N₂, the moles of H₂ (g) required would be

But we have only 4.96 × 10³ mol H₂. Hence, dihydrogen is the limiting reagent in this case. So NH₂(g) would be formed only from that amount of available dihydrogen i.e., 4.96 × 10³ mol
Since 3 mol H₂(g) gives 2 mol NH₃(g)

3.30 × 10³ mol NH₃ (g) is obtained.

If they are to be converted to grams, it is done as follows:
1 mol NH₃ (g) = 17.0 g NH₃(g)

= 3.30 × 10³ × 17 g NH₃ (g)
= 56.1 × 10³ g NH₃
= 56.1 kg NH₃

Question 6.
A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the mass per cent of the solute. [TS Mar. ’19]
Answer:
Mass per cent of

Question 7.
Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in enough water to form 250 mL of the solution. [Mar. ’18 (AP)]
Answer:
Since molarity (M)

Note that molarity of a solution depends upon temperature because volume of a solution is temperature dependent.

Question 8.
The density of 3 M solution of NaCl is 1.25 g mL^-1. Calculate molality of the solution.
Answer:
M = 3 mol L^-1
Mass of NaCl
in 1 L solution = 3 × 58.5 = 175.5 g
Mass of 1L solution = 1000 × 1.25 = 1250 g
(since density = 1.25 g mL^-1)
Mass of water in solution = 1250 – 175.5
= 1074.5 g= 1.0745 kg.

Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution of known higher concentration. The solution of higher concentration is also known as stock solution. Note that molality of a solution does not change with temperature since mass remains unaffected with temperature.

Question 9.
Calculate the normality of oxalic acid so-lutions containing 6.3g of H₂C₂O₄.2H₂O in 500 ml of solutions.
Answer:
Weight of solute = 6.3 g

Question 10.
Calculate the mass of Na₂CO₃ required to prepare 250 ml of 0.5 N solution.
Answer:
Normality of required solution = 0.5 N
Volume of required solution = 250 ml