What is the percent yield in a reaction between 42.6 g O2 and 49.2 g Al if 72.4 g of Al2O3 is produced?

229%

Explanation:

The equation of the reaction is;

4Al(s) + 3O2(g) > 2Al2O3(s)

We must first determine the limiting reactant;

Number of moles of Al2O3 produced = mass/molar mass = 72.4g/101.96 g/mol = 0.71 moles

For Al

Number of moles reacted = mass/molar mass = 49.2g/27 g/mol = 1.8 moles

If 4 moles of Al yields 0.71 moles of Al2O3

1.8 moles of Al will yield 1.8 × 0.71/4 = 0.32 moles of Al2O3

For O2

Number of moles reacted = mass/molar mass = 42.6g/32g/mol = 1.33 moles

If 3 moles of O2 yields 0.71 moles of Al2O3

1.33 moles of O2 will yield 1.33 × 0.71/3 = 0.31 moles of Al2O3

Oxygen is the limiting reactant.

% yield = actual yield/ theoretical yield × 100/1

% yield = 0.71 moles/0.31 moles × 100

% yield = 229%

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