{"id":13261,"date":"2024-03-16T16:56:48","date_gmt":"2024-03-16T11:26:48","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=13261"},"modified":"2024-03-18T17:42:58","modified_gmt":"2024-03-18T12:12:58","slug":"ts-10th-class-physical-science-important-questions-chapter-2","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-10th-class-physical-science-important-questions-chapter-2\/","title":{"rendered":"TS 10th Class Physical Science Important Questions Chapter 2 Chemical Equations"},"content":{"rendered":"

These TS 10th Class Physical Science Chapter Wise Important Questions<\/a> Chapter 2 Chemical Equations will help the students to improve their time and approach.<\/p>\n

TS 10th Class Physical Science Important Questions Chapter 2 Chemical Equations<\/h2>\n

1 Mark Questions<\/span><\/p>\n

Question 1.
\nWhat are the reactants and products in a chemical reaction.
\nAnswer:
\nThe substances which undergo chemical changes In a reaction are called reactants and the new substances formed are called products.<\/p>\n

Question 2.
\nWhat Is meant by balanced equation?
\nAnswer:
\nA chemical equation Is said to be balanced when the number of atoms of each element is same on both reactant side and product side.<\/p>\n

Question 3.
\nIf you keep an iron piece in solid-state CuSO4 crystals, does it get any reaction? Guess the reason.
\nAnswer:
\nReaction will not takes place if an \u00a1ron piece is placed in solid state CuSO4<\/sub> crystals because there will not exist separate Cu2+<\/sup>, SO4<\/sub>-2<\/sup> ions in CuSO4<\/sub> crystals. In aqueous solution, they exists. So iron cannot displace copper.<\/p>\n

Question 4.
\nBaiancethefollowlng equations.
\n(1) Na + O2<\/sub> \u2192 Na2<\/sub>O
\n(2) H2<\/sub>O2<\/sub> \u2192 H2<\/sub>O + O2<\/sub>
\n(3) Mg(OH)2<\/sub> + HCl \u2192 MgCl2<\/sub> + H2<\/sub>O
\n(4) Fe + O2<\/sub> \u2192 Fe2<\/sub>O3<\/sub>
\nAnswer:
\n(1) 4Na + O2<\/sub> \u2192 2Na2<\/sub>O
\n(2) 2H2<\/sub>O2<\/sub> \u2192 2H2<\/sub>O + O2<\/sub>
\n(3) Mg(OH)2<\/sub>+ 2HCl \u2192 MgCl2<\/sub> + 2H2<\/sub>O
\n(4) 4Fe + 3O2<\/sub> \u2192 2Fe2<\/sub>O3<\/sub><\/p>\n

\"TS<\/p>\n

Question 5.
\nBalance the following equations. (AS1)
\n(1) Al(OH)3<\/sub> \u2192 Al2<\/sub>O3<\/sub> + H2<\/sub>O
\n(2) NH3<\/sub> + CuO \u2192 Cu + N2<\/sub> + H2<\/sub>O
\n(3) Al2<\/sub>(SO4<\/sub>)3<\/sub> + NaOH \u2192 Al(OH)3<\/sub> + Na2<\/sub>SO4<\/sub>
\n(4) HNO3<\/sub> + Ca(OH)2<\/sub> \u2192 Ca(NO3<\/sub>)2<\/sub> + H2<\/sub>O
\n(5) NaOH + H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>SO4<\/sub> + H2<\/sub>O
\n(6) BaCl2<\/sub> + H2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub> + HCl
\nAnswer:
\n(1) 2Al(OH)3<\/sub> \u2192 Al2<\/sub>O3<\/sub> + 3H2<\/sub>O
\n(2) 2NH3<\/sub> + 3CuO \u2192 3Cu + N2<\/sub> + 3H2<\/sub>O
\n(3) Al2<\/sub>(SO4<\/sub>)3<\/sub>+ 6NaOH \u2192 2Al(OH)3<\/sub> + 3Na2<\/sub>SO4<\/sub>
\n(4) 2HNO3<\/sub> + Ca(OH)2<\/sub> \u2192 Ca(NO3<\/sub>)2<\/sub> + 2H2<\/sub>O
\n(5) 2NaOH + H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>SO4<\/sub> + 2H2<\/sub>O
\n(6) BaCl2<\/sub> + H2<\/sub>SO4<\/sub> \u2192 BaSO4<\/sub> + 2HCl<\/p>\n

Question 6.
\nWhat is a chemical equation?
\nAnswer:
\nDescribing a chemical reaction using least possible words or symbols is called a chemical equation.
\nEx : CaO + H2<\/sub>O \u2192 Ca(OH)2<\/sub><\/p>\n

Question 7.
\nMnO2<\/sub> + 4HCl \u2192 MnCl2<\/sub>+ 2H2<\/sub>O + Cl2<\/sub>. In the given equation, name the compound which is oxidized and which is reduced?
\nAnswer:
\nIn the above equation Ha compound is oxidized and MnO2<\/sub> is reduced.<\/p>\n

Question 8.
\nWhat happens If the copper sulphate crystals taken into dry test tube are heated?
\nAnswer:<\/p>\n

    \n
  1. When copper sulphate crystals are heated, water present in crystals is evaporated and the salt turns white.<\/li>\n
  2. Evaporated water appears as droplets on the walls of the test tube.<\/li>\n
  3. Blue-coloured copper sulphate (CuSO4<\/sub>5H2<\/sub>O) is turned into white colour beacuse 5H2<\/sub>O molecules are evaporated from crystals.<\/li>\n<\/ol>\n

    Question 9.
    \nWrite the equation for the chemical decomposition reaction of silver chloride In the presence of sunlight.
    \nAnswer:
    \n\"TS<\/p>\n

    2 Marks Questions<\/span><\/p>\n

    Question 1.
    \nWhat do you observe during chemical change?
    \nAnswer:
    \nDuring chemical change, we observe that<\/p>\n

      \n
    1. The original substances have their characteristic properties. Hence these may be products with different physical states colours and different compositions.<\/li>\n
    2. Chemical changes may be exothermic or endothermic.<\/li>\n
    3. They may form an Insoluble substance known as precipitate.<\/li>\n
    4. There may be gas liberation in a chemical danger.<\/li>\n<\/ol>\n

      Question 2.
      \nHow do you Write a chemical equation?
      \nAnswer:<\/p>\n

        \n
      1. A chemical reaction, written tri the form of formula equation, shows the change of reactants to products by an arrow placed between them.<\/li>\n
      2. The reactants are written on the left side of arrow and the final substances or products are written on the right side of the arrow.<\/li>\n
      3. The arrowhead points towards the products and shows the direction of reaction.<\/li>\n
      4. If there are more than one reactant or product involved in the reaction, they are separated with a plus (+) sign between them.<\/li>\n<\/ol>\n

        \"TS<\/p>\n

        Question 3.
        \nWhat are the effects of oxidation on everyday life?
        \nAnswer:
        \n1. Combustion is the most common example for oxidation reactions.
        \n2. Rising of dough with yeast depends on oxidation of sugars to carbon dioxide and water.
        \n3. Metals are corroded due to oxidation.
        \n4. Bleaching of coloured objects using moist chlorine.
        \nCl + H2<\/sub>O \u2192 HOCl + HCl
        \nHOCl \u2192 HCl + (O)
        \nColoured object + (0) \u2192 Colourless object (Here O\u2019 is known as Nascent oxygen)
        \n5. Sometimes during rainy season the power supply to our homes from the electric pole will be interrupted due to formation of metal oxide layer on the electric wire. This metal oxide is an electrical insulator. On removing the metal oxide layer formed on the wire with sandpaper, supply of electricity can be restored.<\/p>\n

        \"TS<\/p>\n

        Question 4.
        \nWhat are the important characteristics of chemical reactions?
        \nAnswer:
        \nThe important characteristics of chemical reactions are<\/p>\n

          \n
        1. Evolution of a gas<\/li>\n
        2. Formation of a precipitate<\/li>\n
        3. Formation of new substances,<\/li>\n
        4. Change in colour<\/li>\n
        5. Change in temperature<\/li>\n
        6. Change in state<\/li>\n<\/ol>\n

          Question 5.
          \nWrite some chemical reactions occurring in our daily life.
          \nAnswer:<\/p>\n

            \n
          1. Souring of milk<\/li>\n
          2. Formation of curd from milk<\/li>\n
          3. Cooking of food<\/li>\n
          4. Digestion of food in our body<\/li>\n
          5. Fermentation of grapes<\/li>\n
          6. Rusting of iron<\/li>\n
          7. Burning of fuels<\/li>\n
          8. Burning of cooking gas<\/li>\n
          9. Ripening of fruits.<\/li>\n<\/ol>\n

            Question 6.
            \nWhat symbols do we use to indicate the physical state of reactants and products In an equation?
            \nAnswer:<\/p>\n

              \n
            1. Solid state is indicated by the symbol (s)<\/li>\n
            2. Liquid state is indicated by the symbol ( l )<\/li>\n
            3. Gaseous state is indicated by the symbol (g)<\/li>\n
            4. Aqueous solution is indicated by the symbol (aq)<\/li>\n
            5. Precipitate Is Indicated by an arrow with head downwards (\u2193)<\/li>\n<\/ol>\n

              Question 7.
              \nComment on \u201cC(s)<\/sub> + O2(g)<\/sub> \u2192 CO2(g)<\/sub> + Heat\u2019 equation.
              \nAnswer:<\/p>\n

                \n
              1. The burning of carbon In oxygen Is an exothermic reaction because heat is evolved in this reaction.<\/li>\n
              2. An exothermic reaction is indicated by writing + Heat or + Heat energy or just+ Energy on product side of an equation.<\/li>\n<\/ol>\n

                Question 8.
                \nComment on \u2018N2(g)<\/sub> + O2(g)<\/sub> + Heat \u2192 2NO(g)<\/sub>\u201d equation.
                \nAnswer:<\/p>\n

                  \n
                1. The reaction between nitrogen and oxygen to form Nitrogen Monoxide is endothermic reaction because heat is absorbed n this reaction.<\/li>\n
                2. An endothermic reaction is usually indicated by writing + Heat or Just \u2018+ Energy on the reactants side of an equation.<\/li>\n<\/ol>\n

                  Question 9.
                  \n2Cu + O2<\/sub> \u2192 2CuO What information do you get from above equation?
                  \nAnswer:
                  \nThe above equation tells us that.<\/p>\n

                    \n
                  1. Copper reacts with oxygen to form copper oxide.<\/li>\n
                  2. The formula of copper oxide.is, CuO and that of oxygen is<\/li>\n
                  3. 2 moles of copper atoms react with 1 mole of oxygen molecules (O2<\/sub>) to produce 2 moles of copper oxide (CuO).<\/li>\n<\/ol>\n

                    Question 10.
                    \nBalance the following chemical equation and follow the steps involved in balancing a chemical equation.
                    \nCu2<\/sub>S + O2<\/sub> \u2192 Cu2<\/sub>O + SO2<\/sub>
                    \nAnswer:
                    \nStep 1: Write the unbalanced equation using correct chemical formula for all substances
                    \nCu2<\/sub>S + O2<\/sub> \u2192 Cu2<\/sub>O + SO2<\/sub>
                    \nStep 2: Compare no. of atoms of each element on both sides.<\/p>\n\n\n\n\n\n\n
                    Atom<\/td>\nNo. of atoms in L.H.S.<\/td>\nNo. of atoms in R.H.S.<\/td>\n<\/tr>\n
                    Cu<\/td>\n2<\/td>\n2<\/td>\n<\/tr>\n
                    S<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
                    O<\/td>\n2<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                    Balancing Cu, S, O atoms both sides
                    \n2Cu2<\/sub>S + 3O2<\/sub> \u2192 2Cu2<\/sub>O + 2SO2<\/sub>
                    \nThe equation is balanced.<\/p>\n

                    Step 3: Write the co-efficient smallest ratio.
                    \n2Cu2<\/sub>S +3O2<\/sub> \u2192 2Cu2<\/sub>O + 2SO2<\/sub><\/p>\n

                    Step 4: Verify above equation for balancing of atom of each element on both sides.
                    \n2Cu2<\/sub>S + 3O2<\/sub> \u2192 2Cu2<\/sub>O + 2SO2<\/sub><\/p>\n

                    Question 11.
                    \nObserve the following balanced chemical equation and answer the given questions.
                    \nC3<\/sub>H8(8)<\/sub> + 5O2(g)<\/sub> \u2192 3CO2(g)<\/sub> + 4H2<\/sub>O(g)<\/sub>
                    \n(i) How many molecules of Oxygen are involved In this chemical reaction?
                    \n(ii) How many moles of Propane are required to get 20 moles of Water?
                    \nAnswer:
                    \n(i) In this chemical reactIon five molecules of oxygen are involved.
                    \n(ii) Five moles of propane are required to get 20 moles of water.<\/p>\n

                    Question 12.
                    \nWrite the products of given reactions, if any. Give reason.
                    \nFeCl2<\/sub> + Zn \u2192
                    \nZnCl2<\/sub> + Fe \u2192
                    \nAnswer:
                    \nFeCl2<\/sub> + Zn \u2192 ZnCl2<\/sub>+Fe (Displacement reaction)
                    \nZnCl2<\/sub> + Fe \u2192 No reaction. (Low reactive metals cannot displace high reactive metals).<\/p>\n

                    \"TS<\/p>\n

                    Question 13.
                    \nBalance the following chemical equations:
                    \n(i) Na+ H2<\/sub>O \u2192 NaOH+H2<\/sub>
                    \n(ii) K2<\/sub>CO3<\/sub>+HCl \u2192KCl+H2<\/sub>O+CO2<\/sub>
                    \nAnswer:
                    \n(i) 2Na + 2H2<\/sub>O \u2192 2NaOH + H2<\/sub>
                    \n(ii) K2<\/sub>CO3<\/sub> +2HCl \u2192 2KCl + H2<\/sub>O+CO2<\/sub><\/p>\n

                    4 Marks Questions<\/span><\/p>\n

                    Question 1.
                    \nWhat are the materials required for the experiment to show the chemical decomposition of water? Write the procedure of the experiment. Name the products which we get in this reaction.
                    \nAnswer:
                    \nMaterial required for chemical decomposition of water:
                    \nA plastic mug, two carbon rods, two corks, two test tubes, connecting wires, 9V battery, water, and some drops of acid.<\/p>\n

                    Procedure:<\/p>\n

                      \n
                    1. Take a plastic mug, drill holes at the base.<\/li>\n
                    2. Fit two one-holed rubber stoppers in these holes.<\/li>\n
                    3. Insert two carbon electrodes in these rubber stoppers.<\/li>\n
                    4. Connect the electrodes to 9V battery.<\/li>\n
                    5. Fill the mug with water, so that the electrodes are immersed.<\/li>\n
                    6. Add few mops of any acid.<\/li>\n
                    7. Take two test tubes tilled with water and insert them over the two carbon electrodes.<\/li>\n
                    8. Switch or\u2019 the circuit and leave the apparatus undisturbed for some time.<\/li>\n
                    9. We notice the liberation of bubbles at both the electrodes. These bubbles displace the water in the test tubes.<\/li>\n
                    10. After the test tubes are filled with gas, take them out and test the gases with the burning match Stick.<\/li>\n<\/ol>\n

                      The products which we get in this reaction:
                      \nThe gases are condensed oxygen and hydrogen.<\/p>\n

                      Question 2.
                      \nExplain the steps involved in balancing a chemical equation with an example.
                      \n(or)
                      \nWhy should we balance a chemical equation? Take any one chemical equation and explain the procedure of balancing it.
                      \nAnswer:
                      \nA chemical equation in which the number of atoms of different elements on the reactants side are same as those on product side is called a balanced equation.
                      \nSteps involved In balancing a chemical reaction: Let us consider the combustion reaction of Propane.
                      \nStep 1: Write the unbalanced equation using correct chemical formulae for all substances.
                      \nC3<\/sub>H8<\/sub> + O2<\/sub> \u2192 CO2<\/sub> + H2<\/sub>O (Skeleton equation)<\/p>\n

                      Step 2: Compare number of atoms of each element on both sides.
                      \n\"TS<\/p>\n

                      Find the coefficients to balance the equation. In this case, there are 3 carbon atoms on the left side of the equation but only one on the right side. If we add a coefficient of 3 to CO2<\/sub> on the right side the carbon atoms balance.
                      \nC3<\/sub>H8<\/sub> + O2<\/sub> \u2192 3CO2<\/sub> + H2<\/sub>O
                      \nNow, look at the number of hydrogen atoms. There are 8 hydrogen atoms on the left but only 2 on the right side. By adding a coefficient of 4 to the H2<\/sub>O on the right side, the hydrogen atoms get balanced.
                      \nC3<\/sub>H8<\/sub>+ O2<\/sub> \u2192 3CO2<\/sub> + 4H2<\/sub>O
                      \nFinally, look at the number of oxygen atoms. There are 2 on the left side but 10 on the right side. By adding a coefficient of 5 to the 02 on the left side, the oxygen atoms get balanced.
                      \nC3<\/sub>H8<\/sub> + 5O2<\/sub> \u2192 3CO2<\/sub> + 4H2<\/sub>O<\/p>\n

                      Step 3: Make sure the coefficients are reduced to their smallest whole number values. The above equation is already with the coefficients in smallest whole numbers. There is no need to reduce its coefficients. Hence the final equation is
                      \nC3<\/sub>H8<\/sub> + 5O2<\/sub> \u2192 3CO2<\/sub> + 4H2<\/sub>O<\/p>\n

                      Step 4: Check the answer. Count the numbers and kinds of atoms on both sides of the equation to make sure they are the same.<\/p>\n

                      Question 3.
                      \nHow to make a chemical equation more informative?
                      \nAnswer:
                      \nChemical equations can be made more informative by expressing the following characteristics of the reactants and products.
                      \n(i) Physical state
                      \n(ii) Heat exchange (exothermic or endothermic change)
                      \n(iii) Gas evolved (if any)
                      \n(iv) Precipitate formed (If any)
                      \n(i) Expressing the physical state: The physical state of the substances maybe mentioned along with their chemical formulae. The different states i e., gaseous, liquid and solid states are represented by the notations(g), (I) and (s) respectively. If the substance is present as a solution in water, it Is represented as (aq).
                      \nEg : Fe2<\/sub>O3(s)<\/sub> + 2Al(s)<\/sub> \u2192 2Fe(s)<\/sub> +Al2<\/sub>O< sub>3(s)<\/p>\n

                      (ii) Heat exchange: Heat is liberated in exothermic reactions and heat is absorbed in endothermic reactions.
                      \nEg: 1. C(s)<\/sub> + O2(g)<\/sub> \u2192 CO2(g)<\/sub> + Q (exothermic reaction)
                      \n2. N2(g)<\/sub>+ O2(g)<\/sub> \u2192 2NO(g)<\/sub> \u2014 Q (endothermic reaction)
                      \n(or) \"TS<\/p>\n

                      (iii) Gas evolved: If a gas is evolved in a reaction, it is denoted by an upward arrow \u2191or (g)
                      \nEg: Zn(s)<\/sub> + H2<\/sub>SO4(aq)<\/sub> \u2192 ZnSO4(aq)<\/sub> + H2(g)<\/sub> \u2191<\/p>\n

                      (iv) Precipitate formed: If a precipitate is formed in the reaction, it Is denoted by a downward arrow.
                      \nEg : AgNO3(aq)<\/sub> + NaCl(aq)<\/sub> \u2192 AgCl(s)<\/sub> \u2193 + NaNO3(aq)<\/sub><\/p>\n

                      Question 4.
                      \nBalance the following chemical equations.
                      \n(a) Na2<\/sub>SO4<\/sub> + BaCl2<\/sub> \u2192 BaSO4<\/sub> + NaCl
                      \n(b) Al4<\/sub>C3<\/sub> + H2<\/sub>O \u2192 CH4<\/sub> + Al(OH)3<\/sub>
                      \n(c) Pb(NO3<\/sub>)2<\/sub> \u2192 PbO + NO2<\/sub> + O2<\/sub>
                      \n(d) Fe2<\/sub>O3<\/sub> + Al \u2192 Al2<\/sub>O3<\/sub> + Fe
                      \nAnswer:
                      \n(a) Na2<\/sub>SO4<\/sub> + BaCl2<\/sub> \u2192 BaSO4<\/sub> + 2NaCl
                      \n(b) Al4<\/sub>C3<\/sub> + 12H2<\/sub>O \u2192 CH4<\/sub> + 4Al(OH)3<\/sub>
                      \n(c) 2Pb(NO3<\/sub>)2<\/sub> \u2192 2PbO + 4NO2<\/sub> + O2<\/sub>
                      \n(d) Fe2<\/sub>O3<\/sub> + 2Al \u2192 Al2<\/sub>O3<\/sub> + 2Fe<\/p>\n

                      \"TS<\/p>\n

                      Solved Problems<\/span><\/p>\n

                      Question 1.
                      \nAl(s)<\/sub> + Fe2<\/sub>O3(s)<\/sub> \u2192 Al2<\/sub>O3(s)<\/sub> + Fe(s)<\/sub> (atomic masses of Al = 27U, Fe = 56U, and O = 16U)
                      \n\"TS
                      \nSuppose that you are asked to calculate the amount of aluminum, required to get 1120 kg of iron by the above reaction.
                      \nSolution:
                      \nAs per the balanced equation
                      \nAluminium \u2192 Iron
                      \n54g \u2192 112g
                      \nX? \u2192 (1120 x 1000)g
                      \n\u2234 xg = \\(\\frac{(1120 \\times 1000) \\mathrm{g} \\times 54}{112 \\mathrm{~g}} \\) = 10000 x 54g = 540000 g (or) 540 kg.
                      \nTo get 1120 kg of iron we have to use 540 kg of aluminum.<\/p>\n

                      Question 2.
                      \nCalculate the volume, mass, and number of molecules of hydrogen liberated when 230 g of sodium reacts with excess of water at STP. (atomic masses of Na = 23U), O=16U, and H=1U)
                      \nThe balanced equation for the above reaction is,
                      \n\"TS
                      \nSolution:
                      \nAs per the balanced equation
                      \n46g of Na gives 2g of hydro9en.
                      \n230g of Na gives …………………? g of hydrogen.
                      \n\\(\\frac{230 \\mathrm{~g} \\times 2 \\mathrm{~g}}{46 \\mathrm{~g}} \\) = 10g of hydrogen.
                      \nI gram molar mass of any gas at STP ie, standard temperature 273K and standard pressure 1 bar occupies 22.4 liters known as gram molar volume.<\/p>\n

                      2.0 g of hydrogen occupies 22.4 liters at STP.
                      \n10.0 g of hydrogen occupies…… ? litres at STP.
                      \n\\(\\frac{10.0 \\mathrm{~g} \\times 22.4 \\text { litres }}{2.0 \\mathrm{~g}}\\) = 112 litres
                      \n2 g of hydrogen i.e, 1 mole of H2<\/sub> contains 6.02 x 1023<\/sup> (H2<\/sub>) molecules 10 g of hydrogen contain ……………………. ?
                      \n\\(\\frac{10.0 \\mathrm{~g} \\times 6.02 \\times 10^{23} \\text { molecules }}{2.0 \\mathrm{~g}} \\) = 30.10 x 1023<\/sup> molecules = 3.01 \u00d7 1024<\/sup> molecules<\/p>\n

                      Question 3.
                      \nCalculate the volume and No. of molecules of CO2<\/sub> liberated at STP if 50g of CaCO3<\/sub> is treated with dilute hydrochloric acid which contains 7.3 g of dissolved HCI gas.
                      \nSolution:
                      \nThe Chemical equation for the above the relation is
                      \nCaCO3<\/sub>(s) + 2HCl (aq) \u2192 CaCl2(aq)<\/sub> + H2<\/sub>O(l) + CO2(g)<\/sub>
                      \nAs per the stoichiometric equation, 100 g of CaCO3<\/sub> reacts with 73 kg of HCl to liberate 44 g of CO2<\/sub>.
                      \nIn the above problem the amount of CaCO3<\/sub>, taken is 50 gm and HCl available is 7.3 g
                      \n100g of CaCo3<\/sub>, require 73g of HCl and 50 g of CaCo3<\/sub> required 36.6 g of HCl but, only 7.3 g of HCI is available.
                      \nHence the product CO2<\/sub> formed depends only on the amount of HCl which Is in the the least amount but not on the amount of CaCO3<\/sub> which is an excess. The reactant available in less amount Is called limiting reagent as it limits the amount of prouct formed.<\/p>\n

                      Therefore, we can write
                      \n73 g of HCl \u2192 44 g CO2<\/sub>
                      \n7.3g of HCl- ?
                      \n\\(\\frac{7.3 g \\times 44 g}{73 g}\\) = 4.4 g
                      \n44 g of CO2<\/sub> occupies 22.4 L volume at STP
                      \n4.4 g of CO2<\/sub> occupies – ?
                      \n\\(\\frac{4.4 g \\times 22.4 L}{44 g} \\) = 2,24L
                      \n44 g of CO2<\/sub> contain 6.023 \u00d7 1023<\/sup> mol of CO2<\/sub> 4.4 g Contains – ?
                      \n\\(\\frac{4.4 g \\times 6.023 \\times 10^{23}}{44 g}\\) = 6.023 \u00d7 1022<\/sup> mol<\/p>\n

                      \"TS<\/p>\n

                      Question 4.
                      \nWrite the equation for the reaction of Zinc with hydrochloric acid and balance the equation. Find out the number of molecules of hydrogen gas produced In this reaction, when 1 mole of HCl completely reacts at S.T.P. [Gram molar volume is 22.4 liters at S.T.P., Avogadro\u2019s number is 6.023 x1023<\/sup>]
                      \nAnswer:
                      \nZn + HCl \u2192 ZnCl2<\/sub>+H2<\/sub>
                      \nZn(s)<\/sub> + 2HCl(l)<\/sub> \u2192 ZnCl2(s)<\/sub>+H2(g)<\/sub>
                      \nFrom the above equation
                      \n2 moles of HCl produces 1 moIe of H2<\/sub> gas
                      \n1 moIe of HCl produces \\(\\frac{1}{2}\\) mole of H2<\/sub> gas
                      \n1 mole of H2<\/sub> gas at S.T.P Contains 6.023 x 1023<\/sup> molecules
                      \n\\(\\frac{1}{2}\\) mole of H2<\/sub> gas contains = \\(\\frac{6.023 \\times 10^{23}}{2}\\)
                      \n= 3.0115 x 1023<\/sup> molecules<\/p>\n","protected":false},"excerpt":{"rendered":"

                      These TS 10th Class Physical Science Chapter Wise Important Questions Chapter 2 Chemical Equations will help the students to improve their time and approach. TS 10th Class Physical Science Important Questions Chapter 2 Chemical Equations 1 Mark Questions Question 1. What are the reactants and products in a chemical reaction. Answer: The substances which undergo … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[16],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/13261"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=13261"}],"version-history":[{"count":4,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/13261\/revisions"}],"predecessor-version":[{"id":13521,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/13261\/revisions\/13521"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=13261"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=13261"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=13261"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}