Combustion reaction

Question

40% of a liquified petroleum gas is composed of propane and the rest is butane. This mixture is put into a room which has a size of 2m x 3m x 4m. How many grams of this mixture can be burned with the air in this room at STP? (remember that 21% of air by volume is oxygen) 

Answer 1

Volume of room = 2m x 3m x 4m = 24m³ = 24,000 liters

Volume of O₂ = 24,000 L x 0.21 = 5040 L

Moles of O₂ at STP = 5040 L x 1 mole/22.4 L = 225 moles O₂

Combustion reactions:

C₃H₈ + 5O₂ ==> 3CO₂ + 4H₂O

2C₄H₁₀ + 13O₂ ==> 8CO₂ + 10H₂O

It takes 5 moles O₂ for 1 moles C₃H₈ and it takes 6.5 moles O₂ for 1 mole C₄H₁₀, thus...

225 moles O₂/11.5 moles O₂ = x moles gas mixture/2 moles gas mixture

x = 39.13 moles of gas mixture

moles of C3H8 = 39.13 moles x 0.4 = 15.65 moles

moles C₄H₁₀ = 39.13 moles x 0.6 = 23.48 moles

mass C₃H₈ = 15.65 moles x 44 g/mole = 689 g

mass C₄H₁₀ = 23.48 moles x 58 g/mole = 1362 g

Grams of mixture = 689 + 1362 = 2051 g