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Important Question: ICSE 2010 Class 10th (Mole Concept and Stoichiometry B) Gay Lassa's Law. Num. 1. 2H2O 2H2 (g) + O2 (g). If a given experiment results in 2500 cm3 of H2 being produced, what volume of O2 is liberated at the same time under the same conditions of temperature and pressure. Sol : - When 2 vol. of hydrogen is produced the 1 vol. of oxygen is liberated. Therefore, When 2500 cm3 of hydrogen is produced then 1/2 2500 cm3 of oxygen is liberated. Hence, volume of oxygen liberated = 1250 cm3. Num. 2. 4NH3+ 5O2 4NO + 6H2 O. If 27 litres of reactants are consumed, what volume of nitrogen monoxide (Nitric oxide) is produced at the same temperature and pressure. Sol : - Volume of reactants = 4 vol. of ammonia + 5 vol. of oxygen = 9 vol. 9 vol. of reactants produces 4 vol. of Nitric oxide Therefore, 27 vol. of reactants will produce 4/9 27 lit. = 12 litres of Nitric oxide. Num. 3. 4N2O + CH4 CO2 + 2H2 O + 4N2. If all volumes are measured at the same temperature and pressure. Calculate the volume of N2O required to give 150 cm3 of steam. Sol : - 2 vol. of steam is produced by 4 vol. of N 2O 1 vol. will be produced by 4/2 vol. of N 2O 150 cm3 of steam will be produced by 4/2 150 = 300 cm3 of N 2O. Num. 4. What volume of oxygen would be required for the complete combustion of 100 litres of ethane according to the following equation. 2C 2H6 + 7O2 4CO2 + 6H2 O. Sol : - 2 vol. of ethane requires 7 vol. of oxygen 1 vol. of ethane will require 7/2 vol. of oxygen 100 litres of ethane will require 7/2 100 = 350 litres of oxygen. Num .5. What vol. of O2 is required to burn completely a mixture of 22.4 dm3 of CH4 and 11.2 dm3 of H2. The reaction are : CH4 + 2O2 CO2 + 2H2 O ; 2H2 + O2 2H2O. Sol : - CH 4 + 2O2 CO 2 + 2H2 O 1 vol. 2 vol. 1 vol. 1 vol. of CH4 requires 2 vol. of O2 22.4 dm3 of CH4 will require 2 22.4 = 44.8 dm3 -------- (i) 2H 2 + O2 2H 2 O 2 vol. 1 vol. 2 vol. 2 vol. of H 2 requires 1 vol. of O 2 1 vol. of H 2 will require vol. of O 2 11.2 dm3 of H 2 will require 1/2 11.2 = 5.6 dm3 ---------- (ii) Total vol. = 44.8 + 5.6 = 50.4 dm3 . Num. 6. 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows: 2CO + O2 2CO2 . Calculate the volume of oxygen used and carbon dioxide formed in the above reaction. Sol.: - 2CO + O2 2CO 2 2 vol. 1 vol. 2 vol. To burn 2 vol. of CO, 1 vol. of O 2 is required and 2 vol. of CO2 is produced To burn 560 ml of CO, 560/2 = 280 ml of O2 will be required and 560 ml of CO2 will be produced. Hence, vol. of O 2required = 280 ml, vol. of CO2produced = 560 ml.
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