Question

The heat evolved in the combustion of methane is given by the equation :

CH₄ (g) + 2O₂ (g)→ CO₂ (g) + 2H₂O (l), ΔH = – 890.3 kJ mol^–1 

(a) How many grams of methane would be required to produce 445.15 kJ of heat of combustion ? 

(b) How many grams of carbon dioxide would be formed when 445.15 kJ of heat is evolved ? 

(c) What volume of oxygen at STP would be used in the combustion process (a) or (b) ?

Answer 1

(a) From the given equation 890.3 kJ of heat is produced from 1 mole of CH₄, i.e., 12 + 4 = 16 g of CH₄

∴ 445.15 kJ of heat is produced from 8 g of CH₄

(b) From the given equation, when 890.3 kJ of heat is evolved, CO₂ formed = 1 mole = 44 g

∴ When 445.15 kJ of heat is evolved, CO₂ formed = 22 g

(c) From the equation, O₂ used in the production of 890.3 kJ of heat = 2 moles = 2 × 22.4 litres at STP = 44.8 litres at STP

Hence, O₂ used in the production of 445.15 kJ of heat = 22.4 litres at STP.