Question

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

2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)

If 378 mol of octane combusts, what volume of carbon dioxide is produced at 13.0 ∘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

378 mole of octane will produce

= 378 mole of octane *(16 mole of CO2 / 2 mole of octane)

= 3024 mole of CO2

Therefore, the number of moles of CO2 = 3024 mole

PV = nRT

where, P = pressure = 0.995 atm

V = volume = ?

n = number of moles = 3024 mole

R = Gas constant

T = temperature = 13.0 + 273 = 286 K

0.995 * V = 3024 * 0.0821 * 286

0.995 * V = 71005

V = 71005 / 0.995 = 71362 L

Therefore, the volume of CO2 produced would be 71362 L