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

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### 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.
Transcribed Image Text:### 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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