When 13.2 mol of Na reacts with excess water according to the equation 2 Na + 2 H_2O → 2 NaOH + H_2, what mass of hydrogen gas (H_2) is produced?

• A. 26.4 mol × 2.02 g/mol = 53.3 g
• B. 13.2 mol × 1 mol H_2 / 2 mol Na × 2.02 g/mol = 13.3 g
• C. 26.4 g
• D. 13.2 g

Answer: (B)

Verified step by step guidance

1. Identify the balanced chemical equation: \$2 \text{Na} + 2 \text{H}_2\text{O} \rightarrow 2 \text{NaOH} + \text{H}_2$.
2. Determine the mole ratio between sodium (Na) and hydrogen gas (H_2) from the balanced equation. For every 2 moles of Na, 1 mole of H_2 is produced.
3. Calculate the moles of hydrogen gas produced using the mole ratio: \(\text{moles of } H_2 = 13.2 \text{ mol Na} \times \frac{1 \text{ mol } H_2}{2 \text{ mol Na}}\).
4. Find the molar mass of hydrogen gas (H_2), which is approximately 2.02 g/mol.
5. Calculate the mass of hydrogen gas produced by multiplying the moles of H_2 by its molar mass: \(\text{mass of } H_2 = \text{moles of } H_2 \times 2.02 \text{ g/mol}\).