The volume of \( \text{NO}_2 \) gas collected over water at 25.0 °C is approximately 24.49 liters.
To calculate the volume of \( \text{NO}_2 \) gas collected over water, we can follow these steps:
1. **Find the pressure of \( \text{NO}_2 \) (partial pressure):**
\[ P_{\text{NO}_2} = P_{\text{total}} - P_{\text{H}_2\text{O}} \]
\[ P_{\text{NO}_2} = 726.0 \, \text{mmHg} - 23.8 \, \text{mmHg} \]
\[ P_{\text{NO}_2} = 702.2 \, \text{mmHg} \]
2. **Convert pressure to atm:**
\[ P_{\text{NO}_2} = 702.2 \, \text{mmHg} \times \left(\frac{1 \, \text{atm}}{760 \, \text{mmHg}}\right) \]
\[ P_{\text{NO}_2} \approx 0.924 \, \text{atm} \]
3. **Use the Ideal Gas Law to find moles of \( \text{NO}_2 \):**
\[ PV = nRT \]
\[ n = \frac{PV}{RT} \]
\[ n = \frac{(0.924 \, \text{atm})(V)}{(0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K})(298.15 \, \text{K})} \]
4. **Convert grams of copper to moles:**
\[ \text{moles of Cu} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{moles of Cu} = \frac{5.79 \, \text{g}}{63.55 \, \text{g/mol}} \]
5. **Use the balanced chemical equation to find the mole ratio of \( \text{NO}_2 \):**
\[ \text{Cu} + 4 \text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + 2 \text{H}_2\text{O} + 2 \text{NO}_2 \]
From the balanced equation, the ratio is 1:2 for Cu to \( \text{NO}_2 \).
6. **Calculate moles of \( \text{NO}_2 \):**
\[ \text{moles of } \text{NO}_2 = 2 \times \text{moles of Cu} \]
7. **Substitute \( n \) and solve for \( V \):**
\[ V = \frac{(0.924 \, \text{atm}) \times V}{(0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K})(298.15 \, \text{K})} \]
\[ V = \frac{(0.924 \, \text{atm}) \times V}{24.49} \]
\[ V \approx 0.0377 \times V \]
8. **Solve for \( V \):**
\[ V \approx \frac{0.924}{0.0377} \]
\[ V \approx 24.49 \, \text{L} \]
Therefore, the volume of \( \text{NO}_2 \) gas collected over water at 25.0 °C is approximately 24.49 liters.