**Chemical Reactions and Yield Calculations**1. **Percent Yield Calculation:** In the reaction \( \text{Mg (s)} + \text{I}_2 \text{(s)} \rightarrow \text{MgI}_2 \text{(s)} \), if 10.0 g of Mg reacts with 60.0 g of \( \text{I}_2 \) and 57.14 g of \( \text{MgI}_2 \) is produced, what is the percent yield?2. **Aspirin Production Calculation:** What mass of aspirin (\( \text{C}_9\text{H}_8\text{O}_4 \)) is produced from 88.5 g of \( \text{C}_7\text{H}_6\text{O}_3 \) assuming a 95.0% yield from the reaction below? \( \text{C}_7\text{H}_6\text{O}_3 \text{(s)} + \text{C}_4\text{H}_6\text{O}_3 \text{(s)} \rightarrow \text{C}_9\text{H}_8\text{O}_4 \text{(s)} + \text{HC}_2\text{H}_3\text{O}_2 \text{(aq)} \) **Explanation:**- The first problem involves calculating the percent yield of a reaction by comparing the actual yield (57.14 g of \( \text{MgI}_2 \)) to the theoretical yield obtained from stoichiometry. - The second problem involves determining the mass of aspirin produced, factoring in a 95.0% yield, when starting with 88.5 g of a reactant \( \text{C}_7\text{H}_6\text{O}_3 \). This requires stoichiometric calculations based on the balanced chemical equation provided.

Given, Mass of Mg = 10.0 g Mass of I2 = 60.0 g Actual yield of MgI2 = 57.14 g To find the limiting reactant ( reactant that gets consumed first ), • Convert the above mass of reactants to moles No.of moles = Mass / Molar mass • Divide the moles with their respective stoichiometric coefficient. • The one with a lowest value will be the limiting reactant. Molar mass of Mg = 24.3 g/mol No.of moles of Mg = 10 g / 24.3 g/mol = 0.4115 mol Molar mass of I2 = 2 × 127 g/mol = 254 g/mol No.of moles of I2 = 60 g / 254 g/mol = 0.2362 mol On dividing with their coefficients, Mg = 0.4115 / 1 = 0.4115 I2 = 0.2362 / 1 = 0.2362 Since I2 has the lowest value, it will be the limiting reagent. From the balanced equation, Mg(s) + I2(s) -----> MgI2(s) 1 mol of I2 produces 1 mol of MgI2. Thus, 0.2362 mol of I2 should produce 0.2362 mol MgI2. Molar mass of MgI2 = ( 24+ 254 ) g/mol = 278 g/mol Mass of MgI2 = No.of moles × Molar mass = 0.2362 mol × 278 g/mol = 65.66 g Thus, the theoretical yield = 65.66 g Actual yield = 57.14 g Theoretical yield = 65.66 g % yield = ( Actual yield / theoretical yield ) × 100 = ( 57.14 g / 65.66 g ) × 100 = 0.8702 × 100 = 87.02 % Thus, the percentage yield is 87.02 %.

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In this reaction: Mg (s) + I2 (s) → Mgl2 (s), if 10.0 g of Mg reacts with 60.0 g of l2, and 57.14 g of Mgl, form, what is the percent vield?

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

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Stoichiometry

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**Chemical Reactions and Yield Calculations**

1. **Percent Yield Calculation:**

In the reaction \( \text{Mg (s)} + \text{I}_2 \text{(s)} \rightarrow \text{MgI}_2 \text{(s)} \), if 10.0 g of Mg reacts with 60.0 g of \( \text{I}_2 \) and 57.14 g of \( \text{MgI}_2 \) is produced, what is the percent yield?

2. **Aspirin Production Calculation:**

What mass of aspirin (\( \text{C}_9\text{H}_8\text{O}_4 \)) is produced from 88.5 g of \( \text{C}_7\text{H}_6\text{O}_3 \) assuming a 95.0% yield from the reaction below?

\( \text{C}_7\text{H}_6\text{O}_3 \text{(s)} + \text{C}_4\text{H}_6\text{O}_3 \text{(s)} \rightarrow \text{C}_9\text{H}_8\text{O}_4 \text{(s)} + \text{HC}_2\text{H}_3\text{O}_2 \text{(aq)} \)

**Explanation:**

- The first problem involves calculating the percent yield of a reaction by comparing the actual yield (57.14 g of \( \text{MgI}_2 \)) to the theoretical yield obtained from stoichiometry.

- The second problem involves determining the mass of aspirin produced, factoring in a 95.0% yield, when starting with 88.5 g of a reactant \( \text{C}_7\text{H}_6\text{O}_3 \). This requires stoichiometric calculations based on the balanced chemical equation provided.

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