I think the easiest way to solve this is to calculate the amount of CO2 from EACH hydrocarbon (as if the hydrocarbon was pure), then see what contribution came from that particular hydrocarbon, as follows:
hexane is C6H14 = 86 and 31.45%
heptane is C7H16 = 100 and 16.15%
octane is C8H18 = 114 and 100-31.45-16.15 = 52.40
2C6H14 + 19O2 ==> 12CO2 + 14H2O
mols C6H14 = g/molar mass = 13,000/86 = approximately 151 (this is a close estimate. You should recalculate ALl on this post to get a better number) since all of the calculations that follow are close estimates.
mols CO2 formed = 151 mols C6H14 x (12 mols CO2/2 mols(C6H14) = about 907 and 907 x 31.45% = about 28,800 mols and that x 44 g/mol = about 12,500 g CO2 from the hexane.
Now you do the same thing for heptane and octane but the combustion equation for heptane and octane will be different. Then you add the grams from each to find the total grams CO2 produced. Post your work if you get stuck.
A liquid fuel mixture contains 31.45 % hexane (C6H14), 16.15 % heptane (C7H16), and the rest octane (C8H18).
What maximum mass of carbon dioxide is produced by the complete combustion of 13.0 kg of this fuel mixture?
2 answers
Please how did u get the 86?
1 answer