Question

A piece of sodium metal reacts completely with water as follows: 2Na(s) + 2H2O(l) ⟶ 2NaOH(aq) + H2(g) The hydrogen gas generated is collected over water at 25.0°C. The volume of the gas is 246 mL measured at 1.00 atm. Calculate the number of grams of sodium used in the reaction. (Vapor pressure of water at 25°C = 0.0313 atm.)

Answer 1

Answer:

$\large \boxed{\text{0.449 g}}$

Explanation:

1. Gather all the information in one place

M_r:   22.99

           2Na + 2H₂O ⟶ 2NaOH + H₂

$\, p_{\text{tot}} = \quad \text{1.00 atm}\\p_{\text{H2O}} = \text{0.0313 atm}$

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

$p_{\text{tot}} = p_{\text{H2}} + p_{\text{H2O}}\\\text{1.00 atm} =p_{\text{H2}} +\text{0.0313 atm}\\p_{\text{H2}} = \text{0.9687 atm}$

(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

$\begin{array}{rcl}\text{0.9687 atm}\times \text{0.246 L} & = & n \times 0.082 06 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{298.15 K}\\0.2383 & = & 24.47n \text{ mol}^{-1}\\\\n & = & \dfrac{0.2383}{24.47\text{ mol}^{-1}}\\\\& = & 0.009740 \text{ mol}\\\end{array}$

3. Moles of Na

The molar ratio is 2 mol Na: 1 mol H₂

$\text{Moles of Na} =\text{0.009 740 mol H}_{2} \times \dfrac{\text{2 mol Na}}{\text{1 mol H}_{2}} = \text{0.019 48 mol Na}$

4. Mass of Na

$\text{Mass of Na} = \text{0.019 48 mol Na} \times \dfrac{\text{22.99 g Na}}{\text{1 mol Na}} = \text{0.449 g Na}\\\\\text{The mass of Na used was \large \boxed{\textbf{0.449 g}} }$