### Iron and Chlorine Reaction: Percent Yield Calculation**Problem Statement:**Iron metal reacts with chlorine gas, giving iron(III) chloride. The balanced chemical equation for this reaction is:\[ 2 \, \text{Fe} (s) + 3 \, \text{Cl}_2 (g) \rightarrow 2 \, \text{FeCl}_3 (s) \]**Question:**If only 16.1 g of FeCl\(_3\) is obtained from 32.0 g of iron and excess Cl\(_2\), what is the percent yield?**Solution Framework:**To calculate the percent yield, follow these steps:1. **Determine the Molar Masses:** - Molar mass of Fe (iron): \( 55.85 \, \text{g/mol} \) - Molar mass of FeCl\(_3\): - Fe: \( 55.85 \, \text{g/mol} \) - Cl: \( 35.45 \, \text{g/mol} \) (since there are three Cl atoms, multiply by 3) Therefore, molar mass of FeCl\(_3\) = \( 55.85 + 3 \times 35.45 \) = \( 55.85 + 106.35 \) = \( 162.2 \, \text{g/mol} \)2. **Calculate the Theoretical Yield:** - From the balanced equation: \[ 2 \, \text{Fe} \rightarrow 2 \, \text{FeCl}_3 \] Therefore, the molar ratio of Fe to FeCl\(_3\) is 1:1. - Moles of Fe used: \[ \frac{32.0 \, \text{g Fe}}{55.85 \, \text{g/mol}} \approx 0.573 \, \text{mol Fe} \] - Since the molar ratio is 1:1, the moles of FeCl\(_3\) formed will also be \( 0.573 \, \text{mol} \). - Theoretical mass of FeCl\(_3\): \[ 0.573 \, \text{mol} \times

Percentage yield can be calculated by dividing actual yield by theoretical yield and multiplying by…

Iron metal reacts with chlorine gas giving iron (III) chloride. The balanced chemical equation for this reaction is: 2 Fe(s) + 3 Cl₂ (g) → 2 FeCl3 (s) If only 16.1 g of FeCl3 is obtained from 32.0 g of iron and excess Cl2, what is the percent yield? %

Iron metal reacts with chlorine gas giving iron (III) chloride. The balanced chemical equation for this reaction is: 2 Fe(s) + 3 Cl₂ (g) → 2 FeCl3 (s) If only 16.1 g of FeCl3 is obtained from 32.0 g of iron and excess Cl2, what is the percent yield? %

Chemistry

10th Edition

ISBN:9781305957404

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

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

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Stoichiometry

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Question

### Iron and Chlorine Reaction: Percent Yield Calculation

**Problem Statement:**
Iron metal reacts with chlorine gas, giving iron(III) chloride. The balanced chemical equation for this reaction is:
\[ 2 \, \text{Fe} (s) + 3 \, \text{Cl}_2 (g) \rightarrow 2 \, \text{FeCl}_3 (s) \]

**Question:**
If only 16.1 g of FeCl\(_3\) is obtained from 32.0 g of iron and excess Cl\(_2\), what is the percent yield?

**Solution Framework:**

To calculate the percent yield, follow these steps:

1. **Determine the Molar Masses:**
- Molar mass of Fe (iron): \( 55.85 \, \text{g/mol} \)
- Molar mass of FeCl\(_3\):
- Fe: \( 55.85 \, \text{g/mol} \)
- Cl: \( 35.45 \, \text{g/mol} \) (since there are three Cl atoms, multiply by 3)

Therefore, molar mass of FeCl\(_3\) = \( 55.85 + 3 \times 35.45 \)
= \( 55.85 + 106.35 \)
= \( 162.2 \, \text{g/mol} \)

2. **Calculate the Theoretical Yield:**
- From the balanced equation:
\[ 2 \, \text{Fe} \rightarrow 2 \, \text{FeCl}_3 \]
Therefore, the molar ratio of Fe to FeCl\(_3\) is 1:1.

- Moles of Fe used:
\[ \frac{32.0 \, \text{g Fe}}{55.85 \, \text{g/mol}} \approx 0.573 \, \text{mol Fe} \]

- Since the molar ratio is 1:1, the moles of FeCl\(_3\) formed will also be \( 0.573 \, \text{mol} \).

- Theoretical mass of FeCl\(_3\):
\[ 0.573 \, \text{mol} \times

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