# Question #5564b

Apr 12, 2017

The reaction will require 0.373 mol of oxygen.

#### Explanation:

First, we need to write a balanced equation for this combustion:

${C}_{3} {H}_{8} + 5 {O}_{2} \rightarrow 3 C {O}_{2} + 4 {H}_{2} O$

This equation assumes complete combustion has occured, meaning that carbon dioxide is one of the products.

Looking at the equation, we see that each one mole of propane $\left({C}_{3} {H}_{8}\right)$ that burns will require five moles of oxygen.

To solve the problem, first convert the mass of propane to moles:

$3.28 g \div 44 \frac{g}{\text{mol}} = 0.0745$ mol of propane

Next, multiply this amount by 5 as given in the proportions set out in the equation:

$0.0745 \times 5 = 0.373 \text{mol of } {O}_{2}$ required.