Question
How many milliliters of 0.126 M H2O4 are needed to neutralize 0.230 g of NaH?
Express the volume in milliliters to three significant digits.
Express the volume in milliliters to three significant digits.
Answers
GPT-4o mini
To determine how many milliliters of 0.126 M \( \text{H}_2\text{O}_4 \) (which I assume is intended to refer to sulfuric acid \( \text{H}_2\text{SO}_4 \)) are needed to neutralize a given mass of sodium hydride \( \text{NaH} \), we can follow these steps:
1. **Calculate moles of NaH**:
- The molar mass of NaH (sodium hydride) is approximately \( 22.99 \, \text{g/mol (Na)} + 1.01 \, \text{g/mol (H)} = 24.00 \, \text{g/mol} \).
- The number of moles of \( \text{NaH} \) can be calculated using the formula:
\[
\text{moles of NaH} = \frac{\text{mass of NaH}}{\text{molar mass of NaH}} = \frac{0.230 \, \text{g}}{24.00 \, \text{g/mol}} \approx 0.00958 \, \text{mol}
\]
2. **Determine moles of H2SO4 needed for neutralization**:
- The reaction between sodium hydride and sulfuric acid can be represented as:
\[
\text{NaH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2
\]
- This indicates that one mole of \( \text{NaH} \) reacts with one mole of \( \text{H}_2\text{SO}_4 \). Therefore, the moles of \( \text{H}_2\text{SO}_4 \) required is equal to the moles of \( \text{NaH} \):
\[
\text{moles of H}_2\text{SO}_4 = 0.00958 \, \text{mol}
\]
3. **Calculate the volume of H2SO4 solution needed**:
- From the concentration formula \( M = \frac{\text{moles}}{\text{volume (L)}} \), we can rearrange it to find volume:
\[
\text{Volume (L)} = \frac{\text{moles}}{M}
\]
- Plugging in the values:
\[
\text{Volume (L)} = \frac{0.00958 \, \text{mol}}{0.126 \, \text{mol/L}} \approx 0.0760 \, \text{L}
\]
- To convert to milliliters:
\[
\text{Volume (mL)} = 0.0760 \, \text{L} \times 1000 = 76.0 \, \text{mL}
\]
Thus, the volume of \( 0.126 \, M \, \text{H}_2\text{SO}_4 \) needed to neutralize \( 0.230 \, \text{g} \) of \( \text{NaH} \) is **76.0 mL**.
1. **Calculate moles of NaH**:
- The molar mass of NaH (sodium hydride) is approximately \( 22.99 \, \text{g/mol (Na)} + 1.01 \, \text{g/mol (H)} = 24.00 \, \text{g/mol} \).
- The number of moles of \( \text{NaH} \) can be calculated using the formula:
\[
\text{moles of NaH} = \frac{\text{mass of NaH}}{\text{molar mass of NaH}} = \frac{0.230 \, \text{g}}{24.00 \, \text{g/mol}} \approx 0.00958 \, \text{mol}
\]
2. **Determine moles of H2SO4 needed for neutralization**:
- The reaction between sodium hydride and sulfuric acid can be represented as:
\[
\text{NaH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2
\]
- This indicates that one mole of \( \text{NaH} \) reacts with one mole of \( \text{H}_2\text{SO}_4 \). Therefore, the moles of \( \text{H}_2\text{SO}_4 \) required is equal to the moles of \( \text{NaH} \):
\[
\text{moles of H}_2\text{SO}_4 = 0.00958 \, \text{mol}
\]
3. **Calculate the volume of H2SO4 solution needed**:
- From the concentration formula \( M = \frac{\text{moles}}{\text{volume (L)}} \), we can rearrange it to find volume:
\[
\text{Volume (L)} = \frac{\text{moles}}{M}
\]
- Plugging in the values:
\[
\text{Volume (L)} = \frac{0.00958 \, \text{mol}}{0.126 \, \text{mol/L}} \approx 0.0760 \, \text{L}
\]
- To convert to milliliters:
\[
\text{Volume (mL)} = 0.0760 \, \text{L} \times 1000 = 76.0 \, \text{mL}
\]
Thus, the volume of \( 0.126 \, M \, \text{H}_2\text{SO}_4 \) needed to neutralize \( 0.230 \, \text{g} \) of \( \text{NaH} \) is **76.0 mL**.