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.)

Answer 1

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