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Stoichiometry

AP Chemistry· difficulty 4/5

A 12.50 g sample of aluminum reacts with 55.00 g of chlorine gas to form AlCl3AlCl_3. After the reaction, the actual yield of AlCl3AlCl_3 is 52.10 g. What is the percent yield?

  • A

    68.7%

  • B

    84.4%

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  • C

    75.2%

  • D

    92.1%

Explanation

Moles Al = 12.50/26.98=0.463312.50/26.98 = 0.4633; moles Cl2=55.00/70.90=0.7758Cl_2 = 55.00/70.90 = 0.7758. Reaction: 2Al+3Cl2→2AlCl32Al + 3Cl_2 \rightarrow 2AlCl_3. Al needs 32⋅0.4633=0.6950\tfrac{3}{2}\cdot 0.4633 = 0.6950 mol Cl2Cl_2, and 0.7758 mol is available — so Al is the limiting reagent. Theoretical AlCl3AlCl_3 = 0.4633⋅133.33=61.780.4633 \cdot 133.33 = 61.78 g. Yield = 52.10/61.78=84.3%≈84.4%52.10/61.78 = 84.3\% \approx 84.4\%.

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