### Example Problem: Calculating Percent Yield in a Chemical Reaction**Problem Statement:**Aqueous hydrobromic acid (HBr) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H₂O). If 8.77 g of water is produced from the reaction of 46.9 g of hydrobromic acid and 42.2 g of sodium hydroxide, calculate the percent yield of water.**Instructions:**Be sure your answer has the correct number of significant digits in it.---**Step-by-Step Solution**:1. **Balanced Chemical Equation**: \[ \text{HBr} (\text{aq}) + \text{NaOH} (\text{s}) \rightarrow \text{NaBr} (\text{aq}) + \text{H}_2\text{O} (\text{l}) \]2. **Identify the Molar Masses**: - HBr: \( \approx 80.91 \, \text{g/mol} \) - NaOH: \( \approx 40.00 \, \text{g/mol} \) - \( \text{H}_2\text{O} \): \( \approx 18.02 \, \text{g/mol} \)3. **Determine the Limiting Reactant**: - Moles of HBr: \( \frac{46.9 \, \text{g}}{80.91 \, \text{g/mol}} \approx 0.579 \, \text{mol} \) - Moles of NaOH: \( \frac{42.2 \, \text{g}}{40.00 \, \text{g/mol}} \approx 1.055 \, \text{mol} \) The limiting reactant is HBr as it has fewer moles compared to NaOH.4. **Calculate Theoretical Yield of \( \text{H}_2\text{O} \)**: The reaction is 1:1, so the moles of \( \text{H}_2\text{O} \) produced will be the same as the moles of HBr used. - Moles of \( \text{H}_2\text{O} \) produced: \( 0.

In this question, we want to determine the theoretical yield of water. You can see details Solution…

Aqueous hydrobromic acid (HBr) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H,). If 8.77 g of water is produced from the reaction of 46.9 g of hydrobromic acid and 42.2 g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it. dlo

Aqueous hydrobromic acid (HBr) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H,). If 8.77 g of water is produced from the reaction of 46.9 g of hydrobromic acid and 42.2 g of sodium hydroxide, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it. dlo

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

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Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

Question

### Example Problem: Calculating Percent Yield in a Chemical Reaction

**Problem Statement:**

Aqueous hydrobromic acid (HBr) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H₂O). If 8.77 g of water is produced from the reaction of 46.9 g of hydrobromic acid and 42.2 g of sodium hydroxide, calculate the percent yield of water.

**Instructions:**

Be sure your answer has the correct number of significant digits in it.

---

**Step-by-Step Solution**:

1. **Balanced Chemical Equation**:
\[
\text{HBr} (\text{aq}) + \text{NaOH} (\text{s}) \rightarrow \text{NaBr} (\text{aq}) + \text{H}_2\text{O} (\text{l})
\]

2. **Identify the Molar Masses**:
- HBr: \( \approx 80.91 \, \text{g/mol} \)
- NaOH: \( \approx 40.00 \, \text{g/mol} \)
- \( \text{H}_2\text{O} \): \( \approx 18.02 \, \text{g/mol} \)

3. **Determine the Limiting Reactant**:
- Moles of HBr: \( \frac{46.9 \, \text{g}}{80.91 \, \text{g/mol}} \approx 0.579 \, \text{mol} \)
- Moles of NaOH: \( \frac{42.2 \, \text{g}}{40.00 \, \text{g/mol}} \approx 1.055 \, \text{mol} \)

The limiting reactant is HBr as it has fewer moles compared to NaOH.

4. **Calculate Theoretical Yield of \( \text{H}_2\text{O} \)**:
The reaction is 1:1, so the moles of \( \text{H}_2\text{O} \) produced will be the same as the moles of HBr used.
- Moles of \( \text{H}_2\text{O} \) produced: \( 0.

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