Question

Solid sodium reacts with liquid water to form hydrogen gas according to the equation 2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g) What is the pressure of hydrogen gas in the 20.0 L headspace of a reactor vessel when 3.34 kg sodium is reacted with excess water at 50.0°C?

Answer 1

Answer:

96.2 atm

Explanation:

Step 1: Write the balanced equation

2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)

Step 2: Calculate the moles corresponding to 3.34 kg of sodium

The molar mass of Na is 22.99 g/mol.

$3.34 \times 10^{3} g \times \frac{1mol}{22.99g} =145mol$

Step 3: Calculate the moles of hydrogen formed from 145 moles of sodium

The molar ratio of Na to H₂ is 2:1. The moles of H₂ formed are 1/2 × 145 mol = 72.5 mol

Step 4: Convert the temperature to the Kelvin scale

K = °C + 273.15 = 50.0°C + 273.15 = 323.2 K

Step 5: Calculate the pressure exerted by the hydrogen gas

We will use the ideal gas equation.

$P \times V = n \times R \times T\\P = \frac{n \times R \times T}{V} = \frac{72.5mol \times 0.0821atm.L/mol.K \times 323.2K}{20.0L} = 96.2 atm$