Percent Yield Chemistry Example 2

Follow the full solution, then compare it with the other examples linked below.

hard

In the reaction

2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3

10.0

g of Al reacts with excess

\text{Cl}_2

to produce

40.0

g of

\text{AlCl}_3

. Calculate the percent yield.

1
Moles of Al $= \frac{10.0}{26.98} = 0.3706\,\text{mol}$ .
2
Mole ratio: 2 mol Al → 2 mol $\text{AlCl}_3$ . Theoretical moles of $\text{AlCl}_3 = 0.3706\,\text{mol}$ .
3
Theoretical mass $= 0.3706 \times 133.34 = 49.42\,\text{g}$ .
4
Percent yield $= \frac{40.0}{49.42} \times 100\% = 80.9\%$ .

Answer

80.9\%

First calculate the theoretical yield using stoichiometry, then compare with the actual yield. Percent yield below 100% is normal in practice.

About Percent Yield

The ratio of the actual yield obtained in an experiment to the theoretical yield predicted by stoichiometry, expressed as a percentage.

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