Asked by Luke
How many milliliters of O2 are consumed in the complete combustion of a sample of hexane, C6H14, if the reaction produces 325 mL of CO2? Assume all gas volumes are measured at the same temperature and pressure. The reaction is
2C6H14(g) + 19O2(g) -> 12CO2(g) + 14H2O(g)
2C6H14(g) + 19O2(g) -> 12CO2(g) + 14H2O(g)
Answers
Answered by
DrBob222
When it's an all gas sample and the T and P don't change, you may use a shortcut and use volume directly as if they were mols.
2C6H14(g) + 19O2(g) -> 12CO2(g) + 14H2O(g)
325 mL CO2 x (19 mols O2/12 mols CO2) = 325 mL x (19/12) = mL O2 consumed.
2C6H14(g) + 19O2(g) -> 12CO2(g) + 14H2O(g)
325 mL CO2 x (19 mols O2/12 mols CO2) = 325 mL x (19/12) = mL O2 consumed.
Answered by
Luke
Around 514.58?
Answered by
DrBob222
Yes, and if your prof is picky about the number of significant figures you whould round that to 514 mL. You are limited to 3 from the 325 to start.
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