Hydracarbon Combustion

Question

A mixture of hydrocarbons contains 38.6%,

C6H14, 13.7% octane, C8H18, and 47.7% decane, C10⁢H22. The mixture is combusted in an excess of oxygen. What mass of CO2 is generated from the combustion of 112.2 kg of the mixture?

Answer 1

Find moles of each hydrocarbon:

moles C6H14: 112.2 kg x 0.386 = 43.31 kg x 1000 g / kg x 1 mol / 86 g = 503.6 moles

moles C8H18: 112.2 kg x 0.137 = 15.37 kg x 1000 g / kg x 1 mol / 114 g = 134.8 moles

moles C10H22: 112.2 kg x 0.477 = 53.52 kg x 1000 g / kg x 1 mol / 142 g = 376.9 moles

Find moles of carbon (C):

503.6 mols C6H14 x 6 mols C / mol = 3022 mols C

134.8 mols C8H18 x 8 mol C /mol = 1078 mols C

376.9 mols C10H22 x 10 mol C / mol = 3769 mols C

Total moles C = 7869 moles of carbon

Moles of CO2 generated:

7869 moles C x 1 mol CO2 / mole C = 7869 moles CO2

Mass of CO2 generated:

7869 moles CO2 x 44 g / mol = 346,236 g = 346.2 kg CO2