Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O.Suppose 61. g of hydrobromic acid is mixed with 44.1 g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

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Aqueous hydrobromic acid HBr will react with solid sodium hydroxide NaOH to produce aqueous sodium bromide NaBr and liquid water H2O.Suppose 61. g of hydrobromic acid is mixed with 44.1 g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits. 

I keep getting this problem wrong because of the sig. digits, dont round till the end because if it is off by 0.1, it will be incorrect.

### Problem Statement:

Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H2O).

Suppose 61.0 g of hydrobromic acid is mixed with 44.1 g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

### Step-by-Step Solution Guide:

1. **Balanced Chemical Reaction:**
   The chemical reaction can be represented as follows:
   \[ \mathrm{HBr(aq) + NaOH(s) \rightarrow NaBr(aq) + H_2O(l)} \]

2. **Molecular Weights:**
   - Hydrobromic acid (HBr): \( \approx 80.91 \, \text{g/mol} \)
   - Sodium hydroxide (NaOH): \( \approx 40.00 \, \text{g/mol} \)

3. **Moles Calculation:**
   - Moles of HBr: \(\frac{61.0 \, \text{g}}{80.91 \, \text{g/mol}} \approx 0.754 \, \text{mol}\)
   - Moles of NaOH: \(\frac{44.1 \, \text{g}}{40.00 \, \text{g/mol}} \approx 1.103 \, \text{mol}\)

4. **Limiting Reactant Analysis:**
   From the balanced equation, the molar ratio of HBr to NaOH is 1:1.
   - Since \( 0.754 \) moles of HBr will react with \( 0.754 \) moles of NaOH.

5. **Excess Reactant Calculation:**
   - Initial moles of NaOH: \( 1.103 \, \text{mol} \)
   - Consumed moles of NaOH: \( 0.754 \, \text{mol} \)
   - Excess moles of NaOH: \( 1.103 - 0.754 = 0.349 \, \text{mol} \)
   - Excess mass of HBr: \( 0.349 \, \text{mol}
Transcribed Image Text:### Problem Statement: Aqueous hydrobromic acid (HBr) will react with solid sodium hydroxide (NaOH) to produce aqueous sodium bromide (NaBr) and liquid water (H2O). Suppose 61.0 g of hydrobromic acid is mixed with 44.1 g of sodium hydroxide. Calculate the minimum mass of hydrobromic acid that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits. ### Step-by-Step Solution Guide: 1. **Balanced Chemical Reaction:** The chemical reaction can be represented as follows: \[ \mathrm{HBr(aq) + NaOH(s) \rightarrow NaBr(aq) + H_2O(l)} \] 2. **Molecular Weights:** - Hydrobromic acid (HBr): \( \approx 80.91 \, \text{g/mol} \) - Sodium hydroxide (NaOH): \( \approx 40.00 \, \text{g/mol} \) 3. **Moles Calculation:** - Moles of HBr: \(\frac{61.0 \, \text{g}}{80.91 \, \text{g/mol}} \approx 0.754 \, \text{mol}\) - Moles of NaOH: \(\frac{44.1 \, \text{g}}{40.00 \, \text{g/mol}} \approx 1.103 \, \text{mol}\) 4. **Limiting Reactant Analysis:** From the balanced equation, the molar ratio of HBr to NaOH is 1:1. - Since \( 0.754 \) moles of HBr will react with \( 0.754 \) moles of NaOH. 5. **Excess Reactant Calculation:** - Initial moles of NaOH: \( 1.103 \, \text{mol} \) - Consumed moles of NaOH: \( 0.754 \, \text{mol} \) - Excess moles of NaOH: \( 1.103 - 0.754 = 0.349 \, \text{mol} \) - Excess mass of HBr: \( 0.349 \, \text{mol}
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