Some commercial drain cleaners contain two components: sodium hydroxide and aluminum powder. When the mixture is poured down a clogged drain, the following reaction occurs: $$2 \mathrm{NaOH}(a q)+2 \mathrm{Al}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow_{2 \mathrm{NaAl}(\mathrm{OH})_{4}(a q)+3 \mathrm{H}_{2}(g)}$$ The heat generated in this reaction helps melt away obstructions such as grease, and the hydrogen gas released stirs up the solids clogging the drain. Calculate the volume of \(\mathrm{H}_{2}\) formed at STP if \(3.12 \mathrm{~g}\) of \(\mathrm{Al}\) is treated with an excess of \(\mathrm{NaOH}\).

Short Answer

Expert verified
The volume of \(H_{2}\) formed at STP when 3.12 g of \(Al\) is treated with an excess of \(NaOH\) is approximately 3.89 L.

Step by step solution

01

Write down the balanced chemical equation

The balanced chemical equation is already provided: \[2 \mathrm{NaOH}(a q)+2 \mathrm{Al}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow_{2} \mathrm{NaAl}(\mathrm{OH})_{4}(a q)+3 \mathrm{H}_{2}(g)\] This equation tells us that 2 moles of Al produce 3 moles of H2.
02

Convert mass of Al to moles

Using the molar mass of Al which is \(26.98 \, \mathrm{g/mol} \), the given mass of Al (3.12 g) can be converted to moles: \[\frac{3.12 \, \mathrm{g}}{26.98 \, \mathrm{g/mol}} \approx 0.116 \, \mathrm{mol} \, \mathrm{Al}\]
03

Use stoichiometric ratios to find moles of H2

From the balanced equation, 2 moles of Al produce 3 moles of H2. Therefore, the moles of H2 produced from 0.116 mol Al are: \(0.116 \, \mathrm{mol} \, \mathrm{Al} \times \frac{3 \, \mathrm{mol} \, \mathrm{H2}}{2 \, \mathrm{mol} \, \mathrm{Al}} \approx 0.174 \, \mathrm{mol} \, \mathrm{H2}\]
04

Convert moles of H2 to volume at STP

At standard temperature and pressure (STP), 1 mole of any gas occupies a volume of 22.4 L. Therefore, the volume of 0.174 mol H2 is: \(0.174 \, \mathrm{mol}\times 22.4 \, \mathrm{L/mol} \approx 3.89 \, \mathrm{L}\)

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