A liquid fuel mixture contains 29.85 hexane , 16.00 heptane , and the rest octan

Question

A liquid fuel mixture contains 29.85 hexane , 16.00 heptane , and the rest octane .

What maximum mass of carbon dioxide is produced by the complete combustion of 13.0 kg of this fuel mixture?

Explanation / Answer

I am assuming your are giving me percents since you didnt provide units Hexane: 0.2985 *13 = 3.88 g 3.88g / 86.2 g = 0.045 mol Hexane 2C6H14 + 19O2 ---> 14H2O + 12CO2 + energy 0.045 mol Hexane * (12 mol Co2 / 2 mol Hexane) *(44 g Co2 / 1 mol CO2) = 11.88 g CO2 Heptane 0.16*13 = 2.08g 2.08 / 100.2 g = 0.021 mol heptane C7H16 + 11O2? 7CO2 + 8H2O .021 mol heptane*(7 mol CO2 / 1mol Heptane)*(44 g Co2 / 1 mol CO2) = 6.47 g CO2 Octane: (1- .16-.2985)*13 = 7.04 g 7.04 / 114.2 g = 0.062 mol Octane 2C8H18 + 25O2--> 16CO2 + 18H2O 0.062 mol Octane*(16 mol CO2 / 2 mol octane)*(44 g Co2 / 1 mol CO2) = 21.82 21.82+ 6.47+11.88 = 40.17 g CO2

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