Question 0c07e

Feb 7, 2016

The combustion consumes 1070 mL of ${\text{O}}_{2}$.

Explanation:

We don't even have to use moles or the Ideal Gas Law to solve this problem.

We can use Gay-Lussac's Law of Combining Volumes — when gases react to form other gases at the same temperature and pressure, they do so in whole-number ratios.

$\text{2C"_6"H"_14 + "19O"_2("g") → "12CO"_2("g") + "14H"_2"O}$
$\textcolor{w h i t e}{m m m m m} \text{19 mL"color(white)(lmml) "12 mL}$

The balanced equation tells us that 19 mL of ${\text{O}}_{2}$ are required to form 12 mL of ${\text{CO}}_{2}$.

676 color(red)(cancel(color(black)("mL CO"_2))) × ("19 mL O"_2)/(12 color(red)(cancel(color(black)("mL CO"_2)))) = "1070 mL O"_2#