QuestionThe burner in a gas grill mixes 24.0 volumes of air for every one volume of propane ( $\left.\mathrm{C}_{3} \mathrm{H}_{8}\right)$ fuel. Given: like all gases, the volume that propane occupies is directly proportional to the number of moles of it at a given temperature and pressure. Air is $21.0 \%$ (by volume) $\mathrm{O}_{2}$ .
1st attempt
Part 1 (1 point)
How many moles of propane are needed to completely use up the oxygen in 24.0 moles of air? $\square$ mol propane

Studdy Solution
Calculate the moles of propane needed to react with the calculated moles of oxygen.
Using the stoichiometry from the balanced equation:
$\text{Moles of propane} = \frac{\text{Moles of } \mathrm{O}_2}{5} = \frac{5.04}{5} = 1.008 \text{ moles of propane}$
The number of moles of propane needed is:
$\boxed{1.008}$

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