QuestionSeptember 20, 2025

Calculate the percent yield of Br_(2) for the chemical reaction 2AgBrarrow 2Ag+Br_(2) if the actual yield is 2.147 g and you start Ewith 6.234 g of AgBr.

Calculate the percent yield of Br_(2) for the chemical reaction 2AgBrarrow 2Ag+Br_(2) if the actual yield is 2.147 g and you start Ewith 6.234 g of AgBr.
Calculate the percent yield of Br_(2) for the chemical reaction 2AgBrarrow 2Ag+Br_(2) if the actual yield is 2.147 g and you start Ewith 6.234 g of AgBr.

Solution
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Answer

85.2\% Explanation 1. Calculate moles of AgBr used n = \frac{6.234\,g}{197.77\,g/mol} = 0.03154\,mol (Molar mass AgBr: 107.87 + 79.90 = 187.77\,g/mol) 2. Find theoretical moles of Br_2 produced 2AgBr \rightarrow Br_2: 2\,mol\,AgBr gives 1\,mol\,Br_2, so 0.03154\,mol\,AgBr gives 0.01577\,mol\,Br_2 3. Calculate theoretical mass of Br_2 m = n \times M = 0.01577\,mol \times 159.80\,g/mol = 2.522\,g 4. Calculate percent yield \text{Percent yield} = \frac{\text{actual}}{\text{theoretical}} \times 100 = \frac{2.147}{2.522} \times 100 = 85.17\%

Explanation

1. Calculate moles of AgBr used<br /> $n = \frac{6.234\,g}{197.77\,g/mol} = 0.03154\,mol$ (Molar mass AgBr: $107.87 + 79.90 = 187.77\,g/mol$)<br />2. Find theoretical moles of $Br_2$ produced<br /> $2AgBr \rightarrow Br_2$: $2\,mol\,AgBr$ gives $1\,mol\,Br_2$, so $0.03154\,mol\,AgBr$ gives $0.01577\,mol\,Br_2$<br />3. Calculate theoretical mass of $Br_2$<br /> $m = n \times M = 0.01577\,mol \times 159.80\,g/mol = 2.522\,g$<br />4. Calculate percent yield<br /> $\text{Percent yield} = \frac{\text{actual}}{\text{theoretical}} \times 100 = \frac{2.147}{2.522} \times 100 = 85.17\%$
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