Answer to Question #108243 in General Chemistry for Lulu

Question

The combustion of octane, C8H18, proceeds according to the reaction

2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)

If 474 mol of octane combusts, what volume of carbon dioxide is produced at 38.0∘C and 0.995 atm?

Answer 1

V₁ - volume of CO2 produced at T₁= 273 K and p₁=1atm.

V₂ - volume of CO2 produced at T₂=38.0∘C=311 K  and p₂=0.995 atm

V₁= 474*22.4=10617.6 L

p₁*V₁/T₁=p₂*V₂/T₂;

V₂=(p_1*V₁*T₂)/(p₂*T₁)

V₂=(1*10617.6 *311)/(0.995*273)=12156.3 L;

Answer: 12156.3 L