How many liters of O2 ,at STP, are required for the complete combustion of 25.7 L of butane, C4H10, at STP? Butane is the fuel in disposable lighters. 2C4H10 + 13O2 → 8CO2 + 10H2O

Answer 1

First, we must write the correctly balanced equation:

2C₄H₁₀ + 13O₂ ==> 8CO₂ +10 H₂O

The ratios of the balanced equation tells us that 2 L of butane requires 13 L of O₂. Using this relationship, we can use dimensional analysis to solve the problem...

25.7 L butane x 13 L O₂ / 2 L butane = 167 L O₂