What is the volume of hydrogen gas at STP released from 2.30 g of sodium metal and water? __Na(s)+ __H₂O(l)→ __NaOH(aq)+ __H₂(g)

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Quizplus · Accepted Answer

The balanced chemical equation for the reaction between sodium metal and water is: $2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g)$. From this, we see that 2 moles of sodium produce 1 mole of hydrogen gas. The molar mass of sodium (Na) is approximately 23 g/mol, so 2.30 g of sodium is $2.30 \, \text{g} \div 23 \, \text{g/mol} = 0.1 \, \text{mol}$. This means 0.1 moles of sodium will produce 0.05 moles of hydrogen gas. At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4 liters, so 0.05 moles of hydrogen gas will occupy $0.05 \, \text{mol} \times 22.4 \, \text{L/mol} = 1.12 \, \text{L}$.

What Is the Volume of Hydrogen Gas at STP Released