Answer:
[tex]\large \boxed{\text{0.449 g}}[/tex]
Explanation:
1. Gather all the information in one place
M_r: 22.99
2Na + 2H₂O ⟶ 2NaOH + H₂
[tex]\, p_{\text{tot}} = \quad \text{1.00 atm}\\p_{\text{H2O}} = \text{0.0313 atm}[/tex]
T = 25.0 °C
V = 246 mL
2. Moles of H₂
To find the moles of hydrogen, we can use the Ideal Gas Law:
pV = nRT
(a) Calculate the partial pressure of the hydrogen
[tex]p_{\text{tot}} = p_{\text{H2}} + p_{\text{H2O}}\\\text{1.00 atm} =p_{\text{H2}} +\text{0.0313 atm}\\p_{\text{H2}} = \text{0.9687 atm}[/tex]
(b) Convert the volume to litres
V = 246 mL = 0.246 L
(c) Convert the temperature to kelvins
T = (25.0 + 273.15) K = 298.15 K
(d) Calculate the moles of hydrogen
[tex]\begin{array}{rcl}\text{0.9687 atm}\times \text{0.246 L} & = & n \times 0.082 06 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{298.15 K}\\0.2383 & = & 24.47n \text{ mol}^{-1}\\\\n & = & \dfrac{0.2383}{24.47\text{ mol}^{-1}}\\\\& = & 0.009740 \text{ mol}\\\end{array}[/tex]
3. Moles of Na
The molar ratio is 2 mol Na: 1 mol H₂
[tex]\text{Moles of Na} =\text{0.009 740 mol H}_{2} \times \dfrac{\text{2 mol Na}}{\text{1 mol H}_{2}} = \text{0.019 48 mol Na}[/tex]
4. Mass of Na
[tex]\text{Mass of Na} = \text{0.019 48 mol Na} \times \dfrac{\text{22.99 g Na}}{\text{1 mol Na}} = \text{0.449 g Na}\\\\\text{The mass of Na used was $\large \boxed{\textbf{0.449 g}}$}[/tex]
