Question

The combustion of octane, C8H18, proceeds according to the reaction shown. 2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l) If 346 mol of octane combusts, what volume of carbon dioxide is produced at 15.0 ∘C and 0.995 atm?

Answer 1

From the balanced equation we can say that

2 mole of octane produces 16 mole of CO2 so

346 mole of octane will produce

= 346 mole of octane *(16 mole of CO2 / 2 mole of octane)

= 2768 mole of CO2

PV = nRT

where, P = pressure = 0.995 atm

V = volume = ?

n = number of moles = 2768 mole

R = Gas constant

T = temperature = 15.0 + 273 = 288 K

0.995 * V = 2768 * 0.0821 * 288

0.995 * V = 65449

V = 65449 / 0.95 = 65778 L

Therefore, the volume of CO2 produced would be 65778 L