QuestionApril 30, 2025

A biologist calculated that 272 g of glucose should have been formed during the reaction of CO_(2) and H_(2)O . What is the percent yield if 32.3 g of glucose are collected at the end of the reaction? 6CO_(2)+6H_(2)Oarrow 6O_(2)+C_(6)H_(12)O_(6)

A biologist calculated that 272 g of glucose should have been formed during the reaction of CO_(2) and H_(2)O . What is the percent yield if 32.3 g of glucose are collected at the end of the reaction? 6CO_(2)+6H_(2)Oarrow 6O_(2)+C_(6)H_(12)O_(6)
A biologist calculated that 272 g of glucose should have been formed during the reaction of CO_(2) and H_(2)O . What is the percent yield if 32.3 g of glucose are collected at the end of the reaction? 6CO_(2)+6H_(2)Oarrow 6O_(2)+C_(6)H_(12)O_(6)

Solution
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Answer

11.88\% Explanation 1. Write the percent yield formula Percent yield is calculated using **\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100**. 2. Substitute the given values Actual yield = 32.3 \, \text{g}, Theoretical yield = 272 \, \text{g}. Substituting into the formula: \text{Percent Yield} = \left(\frac{32.3}{272}\right) \times 100. 3. Perform the calculation \text{Percent Yield} = \left(0.11875\right) \times 100 = 11.875\%.

Explanation

1. Write the percent yield formula<br /> Percent yield is calculated using **$\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100$**.<br /><br />2. Substitute the given values<br /> Actual yield = $32.3 \, \text{g}$, Theoretical yield = $272 \, \text{g}$. Substituting into the formula:<br />$\text{Percent Yield} = \left(\frac{32.3}{272}\right) \times 100$.<br /><br />3. Perform the calculation<br /> $\text{Percent Yield} = \left(0.11875\right) \times 100 = 11.875\%$.
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