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The Mole and Stoichiometry: Question 9

Syllabus 3.3

Structured Extended 6 marks

A technician adds excess dilute sulfuric acid to 5.4 g5.4\ \text{g} of aluminium turnings to generate hydrogen gas for a fuel-cell demonstration.

2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)2\text{Al(s)} + 3\text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Al}_2\text{(SO}_4\text{)}_3\text{(aq)} + 3\text{H}_2\text{(g)}

(ArA_r: Al =27= 27)

(a) Calculate the number of moles of aluminium atoms in 5.4 g5.4\ \text{g}. [2]

(b) Use the balanced equation to calculate the number of moles of hydrogen gas produced. [2]

(c) Calculate the volume of hydrogen gas produced, measured at room temperature and pressure (r.t.p.). [2]

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Worked solution

Part (a): Moles of aluminium

Use moles=massmolar mass\text{moles} = \dfrac{\text{mass}}{\text{molar mass}}, with the molar mass of aluminium atoms equal to Ar(Al)=27 g/molA_r(\text{Al})=27\ \text{g/mol}:

moles of Al=5.427=0.2 mol\text{moles of Al} = \frac{5.4}{27} = 0.2\ \text{mol}

Part (b): Moles of hydrogen gas

The balanced equation, 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)2\text{Al(s)} + 3\text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{Al}_2\text{(SO}_4\text{)}_3\text{(aq)} + 3\text{H}_2\text{(g)}, shows a mole ratio of Al:H2=2:3\text{Al}:\text{H}_2 = 2:3.

So the moles of H2\text{H}_2 are 32\dfrac{3}{2} times the moles of Al\text{Al}:

moles of H2=0.2×32=0.3 mol\text{moles of H}_2 = 0.2\times\frac{3}{2} = 0.3\ \text{mol}

Part (c): Volume of hydrogen gas at r.t.p.

At room temperature and pressure, r.t.p., 1 mol1\ \text{mol} of any gas occupies 24 dm324\ \text{dm}^3:

volume of H2=0.3×24=7.2 dm3\text{volume of H}_2 = 0.3\times24=7.2\ \text{dm}^3

Final answers

  • (a) Moles of Al =0.2 mol=\boxed{0.2}\ \text{mol}
  • (b) Moles of H2=0.3 mol\text{H}_2 = \boxed{0.3}\ \text{mol}
  • (c) Volume of H2\text{H}_2 at r.t.p. =7.2 dm3=\boxed{7.2}\ \text{dm}^3