Question

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

2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)

Calculate the volume of carbon dioxide produced when 562 moles of octane combusts at 28.0 °C and 1.00 bar.

Answer 1

+ → 2 C3H18 2 25 O2 25 16 CO2 + 16 18 H2O 18 no.of moles Calculating number of moles: Number of moles of CO2 = 16 x Number of moles of C8H18 2 Number of moles of CO2 = 8 x Number of moles of C8H18 Number of moles of CO2 = 8 x 562 Number of moles of CO2 = 4496.0 mol Calculating volume: Pressure of CO2 = P = 1 bar = 0.987 atm Volume of CO2 = V = ? number of moles of CO2 = n = 4496.0 mol Gas Constant = R = 0.0821 L.atm/mol.K temperature = T = 28.0 °C - 301 K Ideal gas equation PV = nRT V= v=- nRT P | 4496.0 )( 0.0821 )( 301 0.987 L Volume of CO2 = V = 112569.0