A student reacts 832 mL of 5.43 M Na,CO3 with excess AlClz in the lab, according to the balanced chemical equation below, and finds that she can produce 0.72 mole of Al, (CO3)3. What is the percent yield for her experiment? 3Na,CO3 (aq) + 2 AIC13 (aq) → Al2 (CO3)3(s) + 6N C1(aq) Please express your answer to the second decimal place. Type your answer.
A student reacts 832 mL of 5.43 M Na,CO3 with excess AlClz in the lab, according to the balanced chemical equation below, and finds that she can produce 0.72 mole of Al, (CO3)3. What is the percent yield for her experiment? 3Na,CO3 (aq) + 2 AIC13 (aq) → Al2 (CO3)3(s) + 6N C1(aq) Please express your answer to the second decimal place. Type your answer.
Chemistry
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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
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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![### Calculating Percent Yield in a Chemical Reaction
**Problem Statement:**
A student reacts 832 mL of 5.43 M Na₂CO₃ with excess AlCl₃ in the lab, according to the balanced chemical equation below, and finds that she can produce 0.72 moles of Al₂(CO₃)₃. What is the percent yield for her experiment?
**Balanced Chemical Equation:**
\[3 \text{Na}_2\text{CO}_3 (aq) + 2 \text{AlCl}_3 (aq) \rightarrow \text{Al}_2(\text{CO}_3)_3 (s) + 6 \text{NaCl} (aq) \]
**Instructions:**
Please express your answer to the second decimal place.
**Answer Input:**
[Text Box for Answer]
---
**Explanation:**
To calculate the percent yield, we first determine the theoretical yield based on the given volumes and molarities, then compare it to the actual yield provided.
1. **Find the moles of Na₂CO₃:**
- Volume (V) = 832 mL = 0.832 L (since 1 L = 1000 mL)
- Molarity (M) = 5.43 M
- Moles (n) = Molarity (M) × Volume (V)
\[ n_{\text{Na}_2\text{CO}_3} = 5.43 \, \text{M} \times 0.832 \, \text{L} = 4.51776 \, \text{moles} \]
2. **Determine the limiting reactant:**
Since we have an excess of AlCl₃, the amount of product formed will be determined by the moles of Na₂CO₃.
3. **Use stoichiometry to find the theoretical yield:**
From the balanced equation, 3 moles of Na₂CO₃ produce 1 mole of Al₂(CO₃)₃.
\[ n_{\text{Al}_2(\text{CO}_3)_3} = \frac{4.51776 \, \text{moles of Na}_2\text{CO}_3}{3} \approx 1.50592 \, \text{moles} \]
4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19b343d3-def3-436d-96c5-7f330515e000%2F776bf336-477a-4ef4-bab9-d9adc230678b%2F0xhfqdt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculating Percent Yield in a Chemical Reaction
**Problem Statement:**
A student reacts 832 mL of 5.43 M Na₂CO₃ with excess AlCl₃ in the lab, according to the balanced chemical equation below, and finds that she can produce 0.72 moles of Al₂(CO₃)₃. What is the percent yield for her experiment?
**Balanced Chemical Equation:**
\[3 \text{Na}_2\text{CO}_3 (aq) + 2 \text{AlCl}_3 (aq) \rightarrow \text{Al}_2(\text{CO}_3)_3 (s) + 6 \text{NaCl} (aq) \]
**Instructions:**
Please express your answer to the second decimal place.
**Answer Input:**
[Text Box for Answer]
---
**Explanation:**
To calculate the percent yield, we first determine the theoretical yield based on the given volumes and molarities, then compare it to the actual yield provided.
1. **Find the moles of Na₂CO₃:**
- Volume (V) = 832 mL = 0.832 L (since 1 L = 1000 mL)
- Molarity (M) = 5.43 M
- Moles (n) = Molarity (M) × Volume (V)
\[ n_{\text{Na}_2\text{CO}_3} = 5.43 \, \text{M} \times 0.832 \, \text{L} = 4.51776 \, \text{moles} \]
2. **Determine the limiting reactant:**
Since we have an excess of AlCl₃, the amount of product formed will be determined by the moles of Na₂CO₃.
3. **Use stoichiometry to find the theoretical yield:**
From the balanced equation, 3 moles of Na₂CO₃ produce 1 mole of Al₂(CO₃)₃.
\[ n_{\text{Al}_2(\text{CO}_3)_3} = \frac{4.51776 \, \text{moles of Na}_2\text{CO}_3}{3} \approx 1.50592 \, \text{moles} \]
4
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