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.
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