Stoichiometric Calculations 1. The weight of calcium carbonate required to produce carbon-dioxide that is sufficient for conversion of one 0.1 mole sodium carbonate to sodium bicarbonate is 1) 1gm 2) 10gm 3) 5gm 4)100gm 2. What weight of Oxygen required for the complete combustion of 2.4 grams of Mg? 1) 3.2gm 2) 32gm 3) 1.6gm 4) 16gm 3. How many grams of 80% pure marble stone on calcination can give 14 grams of quick lime? 1) 25gm 2) 20gm 3) 31.25gm 4) 11.2gm 4. The percent weight loss suffered by sodium bicarbonate on strong heating is 1) 36.9% 2) 63.1% 3) 25.68% 4)73.8% 5. 10 grams of a hydrated sodium carbonate, Na2CO3.xH2O, on strong heating suffers a weight loss of 6.29 grams. The value of X 1) 1 2) 6 3) 8 4) 10 6. When 50 grams of sulphur was burnt in air, 4% of the impure residue is left over. Calculate the volume of air required at STP containing 21% of oxygen by volume. 1) 80lit 2) 120lit 3) 160lit 4) 200lit 7. What is the weight of calcium carbonate required for the production of 1 L of carbon dioxide at 270C and 750mm by the action of dilute hydrochloric acid? 1) 2.45gm 2) 4.009gm 3) 8.018gm 4) 40.09gm is

8. The number of grams of NaOH that completely neutralizes 4.9 g of phosphoric acid is 1) 120 2) 24 3) 36 4) 6 9. The volume of oxygen required for the complete combustion of 10lit of methane under the same conditions is 1) 20lit 2) 40lit 3) 10lit 4) 4lit 10. When 20ml of methane and 20ml of oxygen are exploded together and the reaction mixture is cooled to laboratory temperature. The resulting volume of the mixture is 1) 40 ml 2) 20 ml 3) 30 ml 4) 10 ml 11. The volume of CO2 obtained by the complete decomposition of one mole of NaHCO3 at STP is 1) 22.4 L 2) 11.2 L 3) 44.8 L 4) 4.48 L 12. How many liters of CO2 at STP will be formed when 500 ml of 0.1M H2SO4 reacts with excess of Na2CO3? 1) 1.12 2) 11.2 3) 0.224 4) 0.112 13. 1g of Mg is burnt in a vessel containing 0.5g of oxygen. The reactant remaining unreacted is 1) 0.25g of Mg 2) 0.1g of Mg 3) 0.1g of O2 4) 0.75g of Mg 14. A gas mixture contains acetylene and carbon dioxide. 20 lit of this mixture require 20 lit of oxygen under the same conditions for complete combustion. The percentage by volume of carbon dioxide in the mixture is 1) 50% 2) 40% 3) 60% 4) 75%

15. Acetylene can be prepared from calcium carbonate by a series of reactions. The mass of calcium carbonate required to prepare 1mole of acetylene is 1) 200g 2) 100g 3) 250g 4) 320g 16. Sodium carbonate of 92% purity is used in the reaction Na2CO3 + CaCl 2 CaCO3 + 2NaCl. The number of grams of Na2CO3 required to yield 1gm of CaCO3 1) 8.5g 2) 10.5g 3) 11.52g 4) 1.152g 17. Benzene burns in oxygen according to the equation 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O(l). How many liters of oxygen are required at STP for the complete combustion of 39g of liquid benzene? 1) 11.2 2) 22.4 3) 42 4) 84 18. The mass of 80% pure H2SO4 required to completely neutralise 60g of NaOH is nearly 1) 92g 2) 58.8g 3) 73.5g 4) 98g 19. 6 grams of magnesite mineral on heating liberated carbon dioxide which measures 1.12 L at STP. The percentage purity of mineral is 1) 30% 2) 50% 3) 70% 4) 90% 20. The volume of hydrogen at STP required to reduce 7.95 grams of cupric oxide to metal is 1) 1.12lit 2) 11.2lit 3) 0.224lit 4) 2.24lit 21. What is the volume of ammonia obtained starting from 2 L of nitrogen, if the conversion is only 6% efficient in the given conditions? 1) 0.12lit 2) 1.2lit 3) 0.24lit 4) 2. 4lit

22. A mixture of 26ml of H2 and 24ml of Cl 2 is exploded. After reaction the tube contains unreacted 1) 2ml of H2 2) 4ml of H2 3) 2ml of Cl 2 4) 1ml of Cl 2 23. To get 5.6 lit of CO2 at STP weight of CaCO3 to be decomposed is 1) 100g 2) 50g 3) 25g 4) 75g 24. The volume of chlorine that can completely react with 100mg of hydrogen at STP is 1) 1120 ml 2) 1.12 ml 3) 11.2 ml 4) 112 ml 25. 200 ml of pure and dry oxygen is subjected to silent electrical discharge. The volume of ozonised oxygen obtained is 190 ml. What is the percent conversion of oxygen to ozone? 1) 3% 2) 5% 3) 10% 4) 15% 26. The volume of chlorine required for the complete reaction of 10 liters of H2S at STP is 1) 22.4 L 2) 5 lit 3) 10 lit 4) 2.5 lit 27. 6.54g of Zn is dissolved in excess of H2SO4. The weight of ZnSO4 formed and the volume of H2 gas liberated at STP are (molar mass of Zn=65.4 g) 1) 161.4g, 2.24 L 2) 16.14g, 2.24 L 3) 16.14g, 22.4 L 4) 16.14g, 11.2 L 28. 10g of a mixture of CaCO3 and Na2CO3 on ignition suffered a loss in weight of 2.2g. The mass ratio of CaCO3 and Na2CO3 is 1) 1:1 2) 1:1.4 3) 1.4:1 4) 1.75:1 29. A mixture of MgO and Mg weighing 10g is treated with excess of dil HCl. Then 2.24 lit of H2 gas was liberated under STP conditions. The mass of MgO present in the sample is

1) 2.4g 2) 7.6g 3) 8g 4) 2g 30. One liter of a mixture of CO and CO2 is passed over red hot coke when the volume increased to 1.6 L under the same conditions of temperature and pressure. The volume of CO in the original mixture is 1) 400 ml 2) 600 ml 3) 500 ml 4) 800 ml KEY 1) 2 2) 3 3) 3 4) 1 5)4 6) 3 7) 2 8) 4 9) 3 10) 2 11) 2 12) 1 13) 1 14) 3 15) 2 16) 4 17) 4 18) 1 19) 3 20) 4 21) 3 22) 1 23) 3 24) 1 25) 4 26) 3 27) 2 28) 1 29) 2 30) 1 SOLUTIONS 1) The balanced equation for conversion of sodium carbonate to sodium bicarbonate is Na2CO3 + H2O + CO2 2NaHCO3 1 mole of Na2CO3 requires 1 mole i.e. 44gm of CO2 Carbon dioxide required for 0.1 mole of Na2CO3 = 4.4 grams The balanced equation for the decomposition of calcium carbonate is CaCO3 CaO + CO2 1 mole i.e. 44gm of CO2 given by 1 mole i.e. 100 grams of CaCO3 The weight of calcium carbonate to produce the required 4.4 grms carbon dioxide = (4.4X100/44) =10gms 2. Solution: The balanced equation for the combustion of magnesium metal is

2Mg + O2 2MgO 2 moles of Mg = 1 mole of O 2 2X24gm of Mg = 32gm of O 2 2.4 grams of Mg= (2.4X32/2X24) = 1.6grams of O 2 3. Solution: CaCO3 CaO + CO2 1 mole of CaO = 1 mole of CaCO3 56 grams of CaO given by 100 grams of CaCO3 14 grams of CaO given by (14/56)100=25 grams of CaCO3 As purity is 80% i.e. 80 grams of calcium carbonate is present in 100 grams of marble stone 25 grams of calcium carbonate is present in (25X100 /80) =31.25grams of marble stone 4. Solution: On strong heating sodium bicarbonate decomposes and loses carbon dioxide and water. 2NaHCO3 Na2CO3 + H2O + CO2 2 moles of NaHCO3 1 mole of CO2 + 1 mole of H2O (2 84) grams of NaHCO3 shows (44 + 18) =62grams weight loss 100 grams of NaHCO3 shows (100X62/2x84) = 36.9% weight loss. 5. Solution: Hydrated sodium carbonate loses water of hydration on strong heating. Na2CO3.xH2O Na2CO3 + xh2o 1 mole of Na2CO3 x moles of H2O

i.e. 106 grams of Na2CO3 combines with 18x grams of water in the given 10 gm, 3.71 grams of Na2CO3 combines with 6.29 grams of water 106 grams of Na2CO3 combines with 100X6.29/3.71)=180grams of water 18 x = = 180 therefore x=10 6 Sol: Weight of sample of sulphur taken = 50 g Percentage of impurity = 4% Weight of impurity in sample = (50X/100/4) = 2 grams Weight of sulphur in sample = 50 2 = 48 grams Combustion of sulphur gives sulphurdioxide S + O2 SO2 1 mole of S 1 mole of O2 32 grams of S = 22.4 L of O2 at STP 48 grams of S =? Volume of oxygen required at STP = (48/32) 22.4 = 33.6 L 21 L of O2 is present in 100 L of air 33.6 L of O2 present in (33.6/21) X100== 160 lit of air. 7. Sol: Given conditions STP conditions P1 = 750 mm T1 = 300K P2 = 760 mm T2 = 273K V1 = 1 Lit V2 =? From the equation of state, the unknown volume is obtained

P1V 1 T2 750 1 273 T1 P2 300 760 V2 = = = 0.898 L Calcium carbonate on reaction with hydrochloric acid gives carbon dioxide. CaCO3 + 2HCl CaCl2 + H2O + CO2 1 mole of CO2 given by 1 mole of CaCO3 22.4 L of CO2 at STP = 100 grams of CaCO3 0.898 L of CO2 at STP =? The weight of calcium carbonate required = 8. Solution: 3NaOH + H 3 PO 4 Na 3 PO 4 + 3H 2 O = 4.009 grams 3molesi.e 3X40gm of NaOH neutralize 1mole i.e. 98 gm of H 3 PO 4 Wt of NaOH that neutralize 4.9gm of H 3 PO 4 =(4.9X3X40/98)=6gm 9. Solution: CH 4 + 2O 2 CO 2 + 2H 2 O 1mol Methane requires 2mol of Oxygen i.e 1 lit of Methane requires 2lit of Oxygen 10lit of Methane requires =10X2=20lit of Oxygen. 10. Solution: CH 4 + 2O 2 CO 2 + 2H 2 O 100 0.898 22.4 1mol Methane reacts with 2mol of Oxygen to give 1mol of CO 2 i.e. 1 ml of Methane reacts with 2ml of Oxygen to give 1ml of CO 2 10m of Methane reacts with 20ml of Oxygen to give 10ml of CO 2. Vol. of methane left=10ml, vol of CO 2 formed=10ml Total vol =20ml

11. Solution: 2NaHCO3 Na2CO3 + H2O + CO2 2 moles of NaHCO3 1 mole i.e. 22.4 lit of CO2 STP 1 mole of NaHCO3 gives 22.4/2 =11.2 lit of CO2 STP 12. Solution: moles of H2SO4 in solution = MX Vin lit =0.1 X500/1000) =0.05 Na 2 CO 3 + H2SO 4 Na2SO4 + H2O + CO 2 1mole of sulphuric acid gives 1mole i.e. 22.4 lit of CO 2 0.05 moles of sulphuric acid gives 0.05X22.4 = 1.12 lit of CO 2 13. Solution: 2Mg + O 2 2 MgO 2moles i.e. 2X24 gm of Mg reacts with 1mole i.e. 32 gm of Oxygen. Wt of Mg that reacts with 0.5gm of Oxygen =( 0.5X2X24/32)=0.75gm Wt of Mg left +1-0.75 = 0.25gm 14. Solution: CO 2 is not combustable. If vol of C 2 H 2 =x then vol. of CO 2 =20-x 2 C 2 H 2 +5O 2 4 CO 2 +2H 2 O 2 vol. of acetylene reacts with 5 vol. of oxygen 2lit of acetylene reacts with 5 lit.of oxygen Of acetylene reacts with 20 lit.of oxygen Vol. of acetylene = (20X2/5) =8lit Vol. of CO 2 in the mixture= 20-8 =12lit Vol% of Co 2 = (12X100/20) =60% 15. Solution: CaCO 3 +3C CaC 2 +3CO CaC 2 +2H 2 O C 2 H 2 +Ca(OH) 2, as per equations 1mole i.e. 100gm of CaCO 3 gives 1mole of C 2 H 2

16. Solution 100gm of CaCO 3 given by 106 gm of Na2CO 3 1gmof CaCO 3 given by (106/100) =1.06gm of Na2CO 3 Wt of 92% Na 2 CO 3 required = (1.06 X100/92) =1.152gm. 17. Solution: 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l). 2moles i.e. 2X78=156 gm of Benzene requires 15X22.4 lit of oxygen at STP 39 gm of Benzene requires (39X15X22.4/156) = 84 lit of oxygen at STP 18. Solution: 2NaOH + H 2 SO 4 Na 2 SO 4 + 2H 2 O 2moles i.e. 2X40=80gm NaOH is given by 98gm of H 2 SO 4 60gm NaOH is given by (60X98/80) = 73.5gm of H 2 SO 4 Wt.of 80% acid required= (73.5X100/80) = 91.875gm 19. Solution: Magnesite is magnesium carbonate. On heating it liberates carbon dioxide. MgCO3 MgO + CO2 1 mole of CO2 is given by 1 mole of MgCO3 22.4 L of CO2 at STP is given by 84 grams of MgCO3 1.12 L of CO2 at STP =? The weight of pure MgCO3 to be decomposed = (1.12 84/22.4) = 4.2 grams 6 grams mineral has 4.2 grams MgCO3 Percent purity = 100 = 70% 20. Solution: Cupric oxide is reduced to copper on treating with hydrogen CuO + H2 Cu + H2O 1 mole of CuO requires 1 mole of H2

79.5 grams of CuO requires 22.4 L of H2 at STP 7.95 grams of CuO requires (7.95 22.4/79.5) = 2.24 L of hydrogen required at STP 21. Solution: Conversion of nitrogen to ammonia is given as N2 + 3H2 2NH3 1 mole of N2 = 2 moles of NH3 1 volume of N2 = 2 volumes of NH3 Volume of ammonia with 100% efficient conversion of 2 L nitrogen = 4 L Volume of ammonia with 6% efficient conversion of the reaction = (6X4/100) = 0.24 L 22. Solution: H 2 + Cl 2 2HCl 1 vol. of H 2 reacts with 1 vol. of Cl 2 1 ml. of H 2 reacts with 1 ml. of Cl 2? ml. of H 2 reacts with 24 ml. of Cl 2 Vol. of H 2 reacts =24ml. Vol. of H 2 left =26-24= 2ml. 23. Solution: CaCO 3 CaO+ CO 2 1mole i.e. 22.4 lit of CO 2 is given by 1mole i.e. 100 gm CaCO 3 5.6 lit of CO 2 is given by (5.6X100 /22.4) =25 gm CaCO 3 24. Solution: H 2 + Cl 2 2HCl 1 mol. of H 2 reacts with 1 mol. of Cl 2 2 gm i.e. 2000mg of H 2 reacts with 22400 ml. of Cl 2 100mg of H 2 reacts with (100X22400/2000) =1120ml. of Cl 2 25. Solution

3O2 2O3 3 vol 2 vol 200 ml 0 ml (200-3x) ml 2x ml Ozonised oxygen = O2 remaining + O3 obtained = 190 ml = (200 3x + 2x) ml Solving, x = 10 ml Volume of O2 remaining = 200 30 = 170 ml Volume of O2 consumed = 30 ml If 200 ml O2 is taken = 30 ml is converted If 100 ml O2 is taken =? The percentage conversion of oxygen = 30 = 15% 26. Solution: Cl2 + H2S 2HCl + S 1 vol. of chlorine reacts with 1 vol. of H 2 S 1 lit.of chlorine reacts with 1 lit.of H 2 S 10lit.of chlorine reacts with 10lit.of H 2 S 27. Solution: Zn +H 2 SO 4 H 2 +ZnSO 4 1mole i.e. 65.4 gm of Zn gives 1mole i.e. 161.4 gm of ZnSO 4 and 22. 4 lit H 2 at STP 6.54 gm of Zn gives 16.14 gm of ZnSO 4 and 2.24 lit H 2 at STP 28. Solution: Na2CO 3 is thermally stable and does not decompose on ignition but CaCO 3 decompose to give CO 2. CaCO3 CaO + CO2 1mole i.e. 100gm of CaCO 3 gives 44 gm of CO 2.

Wt. of CaCO 3 giving 2.2 gm of CO 2 = (2.2X100/44) =5gm Wt. of CaCO 3 =5gm, Wt. of Na2CO 3 =10-5=5gm. Therefore, ratio of wts=1:1 29. Solution: Among Mg and MgO, only Mg gives Hydrogen gas with acid. Mg + 2HCl MgCl 2 +H 2 1 mole i.e. 24gm of Mg gives 22.4 lit of H 2 at STP. Wt.of Mg giving 2.24 lit of H 2 at STP= (2.24X24/22.4)=2.4gm Wt.of Mg=2.4gm and wt of MgO in the mixture= 10-2.4= 7.6gm. 30. Solution: Among CO and CO 2, only CO 2 reacts with coke. Vol.of CO 2 =X lit then vol. of CO = 1-X CO 2 +C 2CO 1 vol of CO 2 gives 2 vol.of CO X lit of CO 2 gives 2X lit.of CO. Final vol. of mixture= (1-X)+2X= 1+X=1.6 lit i.e. X=0.6lit Therefore volume of CO in the mixture = 1-X= 1-0.6= 0.4lit i.e. 400ml