Question

How many grams of Oxygen are needed to combust 20 grams of propane?

Answer 1

Let’s start by writing a balanced equation of the reaction between oxygen and propane :

C₃H₈ + 5O₂ $\rightarrow$ 3CO₂ + 4H₂O

As you can see, 1 mole of propane needs 5 moles of oxygen. The molar mass of propane is 44.1 g/mol. The number of the moles of propane is then:

$n = \frac{m}{M} = \frac{20 \text{ g}}{44.1 \text{ g/mol}} = 0.454 \text{ mol}$

The number of the moles of oxygen needed is five times higher , or 2.27 mol. The molar mass of oxygen is 32 g/mol. The mass of oxygen needed is :

$m = n \cdot M = 2.27 \text{ (mol)} \cdot 32\text{ (g/mol)} = 72.6 \text{ g}$

Answer: 72.6 g of oxygen are needed to burn 20 g of propane.