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)
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.
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.
To determine the amount of O2 consumed in the combustion of hexane, we need to use the stoichiometry of the balanced chemical equation. The coefficient ratio between C6H14 and O2 tells us the ratio of their volumes.
From the balanced chemical equation:
2C6H14(g) + 19O2(g) -> 12CO2(g) + 14H2O(g)
The coefficient ratio between C6H14 and O2 is 2:19.
The coefficient ratio between O2 and CO2 is also 19:12.
Since we know that 325 mL of CO2 is produced, we can use the ratio to calculate the volume of O2 consumed.
1. Convert the volume of CO2 to the volume of O2:
325 mL CO2 x (19/12) = 513.33 mL O2
Therefore, approximately 513.33 mL of O2 are consumed in the complete combustion of hexane.