Answer to Question #318132 in Physical Chemistry for taqua

Question

Assuming no change in temperature and pressure, calculate the volume of O₂ (in liters) required for the complete combustion of 17.6 L of butane (C₄H₁₀):

2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)

 

Answer 1

2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l);

In this case:

n(O₂) = 13/2 * n(C₄H₁₀);

And,

Volume of the butane: V(C₄H₁₀) = 17.6 L;

Molar volume: V_m = 22.4 L/mol;

n(C₄H₁₀) = V(C₄H₁₀)/V_m = 17.6/22.4 mol;

n(O₂) = 13/2 * n(C₄H₁₀) = 13/2 * 17.6/22.4;

V(O₂) = n(O₂) * V_m = 13/2 * 17.6/22.4 * 22.4 = 13/2 * 17.6 = 114.4 L.

Answer: 114.4 L.