When 2.3 g of Na(s) reacts with excess water, what volume of hydrogen gas (H_2) at STP is produced?

• A. 0.56 L
• B. 1.12 L
• C. 4.48 L
• D. 2.24 L

Answer: (B)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction of sodium (Na) with water (H_2O): \$2\mathrm{Na}(s) + 2\mathrm{H_2O}(l) \rightarrow 2\mathrm{NaOH}(aq) + \mathrm{H_2}(g)$.
2. Calculate the number of moles of sodium reacting using its molar mass: \(\text{moles of Na} = \frac{2.3\,\mathrm{g}}{22.99\,\mathrm{g/mol}}\).
3. Use the stoichiometric ratio from the balanced equation to find moles of hydrogen gas produced. Since 2 moles of Na produce 1 mole of H_2, moles of H_2 = \(\frac{1}{2} \times\) moles of Na.
4. Recall that at STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. Calculate the volume of hydrogen gas produced using: \(\mathrm{Volume\ of\ H_2} = \text{moles of H_2} \times 22.4\,\mathrm{L/mol}\).
5. Combine all the calculations to find the volume of hydrogen gas produced when 2.3 g of sodium reacts with excess water at STP.