### Calculating Percent Yield in a Chemical Reaction**Problem Statement:**In this reaction:\[ \text{Mg} (s) + \text{I}_2 (s) \rightarrow \text{MgI}_2 (s) \]If 10.0 g of Mg reacts with 60.0 g of I2, and 60.36 g of MgI2 forms, what is the percent yield?**Details of Calculation:**1. **Balanced Chemical Equation**: \[ \text{Mg} (s) + \text{I}_2 (s) \rightarrow \text{MgI}_2 (s) \]2. **Molar Mass Calculation**: - Molar mass of \( \text{Mg} \): 24.305 g/mol - Molar mass of \( \text{I}_2 \): 2 × 126.904 g/mol = 253.808 g/mol - Molar mass of \( \text{MgI}_2 \): 24.305 g/mol + 2 × 126.904 g/mol = 278.113 g/mol3. **Stoichiometric Calculations**: - Moles of \( \text{Mg} \): \( \frac{10.0 \, \text{g}}{24.305 \, \text{g/mol}} = 0.4112 \, \text{mol} \) - Moles of \( \text{I}_2 \): \( \frac{60.0 \, \text{g}}{253.808 \, \text{g/mol}} = 0.2364 \, \text{mol} \) Since Mg and I2 react in a 1:1 ratio: - \( \text{Mg} \) is the limiting reagent since its molar quantity is less than that of \( \text{I}_2 \).4. **Theoretical Yield**: - Moles of \( \text{MgI}_2 \) produced: 0.4112 mol (from \( \text{Mg} \)) - Theoretical mass of \( \text{MgI}_2 \): \( 0.4112 \, \text{mol} \times 278.113 \, \text{g

The balanced equation is Mg (s) + I2 (s) MgI2 (s) First, find the limiting reactant by calculating…

Answered: In this reaction: Mg (s) + 1₂ (s) →…

In this reaction: Mg (s) + 1₂ (s) → Mgl₂ (s), if 10.0 g of Mg reacts with 60.0 g of 12, and 60.36 g of Mgl₂ form, what is the percent yield?

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Stoichiometry

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Question

### Calculating Percent Yield in a Chemical Reaction

**Problem Statement:**
In this reaction:
\[ \text{Mg} (s) + \text{I}_2 (s) \rightarrow \text{MgI}_2 (s) \]

If 10.0 g of Mg reacts with 60.0 g of I<sub>2</sub>, and 60.36 g of MgI<sub>2</sub> forms, what is the percent yield?

**Details of Calculation:**

1. **Balanced Chemical Equation**:
\[ \text{Mg} (s) + \text{I}_2 (s) \rightarrow \text{MgI}_2 (s) \]

2. **Molar Mass Calculation**:
- Molar mass of \( \text{Mg} \): 24.305 g/mol
- Molar mass of \( \text{I}_2 \): 2 × 126.904 g/mol = 253.808 g/mol
- Molar mass of \( \text{MgI}_2 \): 24.305 g/mol + 2 × 126.904 g/mol = 278.113 g/mol

3. **Stoichiometric Calculations**:
- Moles of \( \text{Mg} \): \( \frac{10.0 \, \text{g}}{24.305 \, \text{g/mol}} = 0.4112 \, \text{mol} \)
- Moles of \( \text{I}_2 \): \( \frac{60.0 \, \text{g}}{253.808 \, \text{g/mol}} = 0.2364 \, \text{mol} \)

Since Mg and I2 react in a 1:1 ratio:
- \( \text{Mg} \) is the limiting reagent since its molar quantity is less than that of \( \text{I}_2 \).

4. **Theoretical Yield**:
- Moles of \( \text{MgI}_2 \) produced: 0.4112 mol (from \( \text{Mg} \))
- Theoretical mass of \( \text{MgI}_2 \): \( 0.4112 \, \text{mol} \times 278.113 \, \text{g

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