Question

calculate the weight and volume of air required for complete combustion of 4m³ of ch4

Answer 1

The equation for the combustion is :

CH₄(g)  + 2O₂ (g) — >  CO₂ (g) +  2H₂O (g)

Mole ratio is 1 : 2

Moles of CH₄ :

1 mole of a gas at stp = 22.4 litres

1 cubic meter = 1000 litres

Converting 4m³ to litres

4 × 1000 = 4000litres

4000/22.4 = 178.57 moles

Moles of Oxygen = 2 × 178.57

2 × 178.57 = 357.14 moles

Molar mass of Oxygen gas = 32

Mass = 32 × 357.14 moles = 11428.48 grams

Volume :

357.14 × 22.4 = 7999.936 litres

= 7.999936 m³