1.A liquid fuel mixture contains 29.50 % hexane (C6H14), 17.70 % heptane (C7H16)
ID: 707585 • Letter: 1
Question
1.A liquid fuel mixture contains 29.50 % hexane (C6H14), 17.70 % heptane (C7H16), and the rest octane (C8H18). What maximum mass of carbon dioxide is produced by the complete combustion of 10.0 kg of this fuel mixture? 2.Many home barbeques are fueled with propane gas (C3H8).What mass of carbon dioxide (in kg) is produced upon the complete combustion of 18.9 L of propane (approximate contents of one 5-gallon tank)? Assume that the density of the liquid propane in the tank is 0.621 g/mL. (Hint: Begin by writing a balanced equation for the combustion reaction.)
2.What mass of carbon dioxide (in kg) is produced upon the complete combustion of 18.9 L of propane (approximate contents of one 5-gallon tank)? Assume that the density of the liquid propane in the tank is 0.621 g/mL. (Hint: Begin by writing a balanced equation for the combustion reaction.)
Explanation / Answer
Ans 1
Mass of hexane (C6H14) in mixture = 10 x 0.2950
= 2.95 kg
Moles of hexane = mass/molecular weight
= 2.95kg/(86.17kg/kmol)
= 0.03423 kmol
Balanced combustion equation:
2 C6H14 + 19 O2 = 12 CO2 + 14 H2O
Maximum moles of CO2 produced
= 12 kmol CO2 x 0.03423 kmol C6H14 / 2 kmol C6H14
= 0.2054 kmol
Mass of heptane (C7H16) in mixture = 10 x 0.177
= 1.77 kg
Moles of heptane = 1.77/100.2 = 0.01766 kmol
Balanced equation:
C7H16 + 11 O2 = 7 CO2 + 8 H2O
Maximum moles of CO2 produced
= 7 kmol CO2 x 0.01766 kmol C7H16 / 1 kmol C7H16
= 0.1236 kmol
Mass of octane (C8H18) in mixture = 10 x (1-0.177-0.2950)
= 5.28 kg
Moles of octane = 5.28/114.23= 0.04622 kmol
Balanced equation:
2 C8H18 + 25 O2 = 16 CO2 + 18 H2O
Maximum moles of CO2 produced
= 16 kmol CO2 x 0.04622 kmol C8H18 / 2 kmol C8H18
= 0.3698 kmol
Total kmoles of CO2 produced = 0.2054 + 0.1236 + 0.3698
= 0.6988 kmol
Mass of CO2 produced = kmol x molecular weight
= 0.6988 kmol x 44 kg/kmol
= 30.7472 kg
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