The aluminum combustion reaction is described by the following chemical reaction equation:
4Al + 3O2 = 2Al2O3;
Let’s calculate the amount of substance contained in 13.5 grams of aluminum.
M Al = 27 grams / mol;
N Al = 13.5 / 27 = 0.5 mol;
To burn 0.5 mol of metal, 0.5 x 3/4 = 0.375 mol of oxygen is required.
Let’s calculate its volume.
1 mole of ideal gas fills a volume of 22.4 liters under normal conditions.
V O2 = 0.375 x 22.4 = 8.4 liters;
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