Hydrogen collection over water and an alloy that is not pure magnesium
A 0.500 g sample of a magnesium-aluminium alloy is treated with 40.0 mL of 2.00 mol/L hydrochloric acid, which is in excess. The hydrogen produced is collected over water at 22.0 degC. The barometric pressure is 748 mmHg and the volume of gas collected is 593 mL. The vapour pressure of water at 22.0 degC is 19.8 mmHg. The student assumes that only magnesium reacts and reports the magnesium content as the mass of metal that would give this volume of hydrogen. A second 0.500 g sample of the same alloy is treated with excess concentrated sodium hydroxide, which dissolves aluminium but not magnesium; under the same collection conditions the volume of hydrogen is 281 mL. The acid and base are both in excess, and the collected gas is saturated with water vapour. The student must decide whether the first result is physically reasonable and how the second experiment can correct it.
1. Use the first experiment to calculate the apparent mass of magnesium and the percentage of the 0.500 g sample it represents. Is the result physically possible if the alloy contained only magnesium? Show your reasoning.
Read the worked answer
The hydrogen is collected over water, so the partial pressure of hydrogen is
P(H2) = 748 - 19.8 = 728.2 mmHg = 728.2/760 = 0.9582 atm.
Using the ideal gas law with V = 0.593 L and T = 295.15 K:
n(H2) = PV/RT = (0.9582)(0.593)/[(0.08206)(295.15)] = 0.02346 mol.
For Mg + 2HCl -> MgCl2 + H2, one mole of Mg gives one mole of H2, so the apparent mass of Mg is
m(Mg) = 0.02346 x 24.31 = 0.570 g.
This is 0.570/0.500 x 100 = 114% of the sample. A value above 100% is impossible for a pure magnesium sample, so the assumption that only magnesium produces hydrogen must be false. Aluminium also reacts with acid and produces more hydrogen per gram than magnesium does.
2. Use the sodium hydroxide result to calculate the masses of aluminium and magnesium in the 0.500 g sample, and reconcile the two experiments quantitatively.
Read the worked answer
With excess NaOH, aluminium reacts according to
2Al + 2NaOH + 6H2O -> 2NaAl(OH)4 + 3H2,
so n(H2) = 1.5 n(Al). The hydrogen from the NaOH experiment is
n(H2) = (0.9582)(0.281)/[(0.08206)(295.15)] = 0.01112 mol.
Therefore n(Al) = 0.01112/1.5 = 0.007413 mol, and
m(Al) = 0.007413 x 26.98 = 0.200 g.
The magnesium mass is 0.500 - 0.200 = 0.300 g. Checking the acid experiment: Mg gives 0.300/24.31 = 0.01234 mol H2, and Al gives 0.01112 mol H2, so total H2 = 0.02346 mol. This is exactly the amount calculated from 593 mL, so the two experiments are consistent. The first experiment overestimated Mg because Al produces 1.5 mol H2 per mol Al, or 0.0556 mol H2 per gram, whereas Mg produces only 0.0411 mol H2 per gram.
3. Show by calculation that 40.0 mL of 2.00 mol/L HCl is in excess for this sample, and explain how the calculated composition would be affected if the acid were insufficient.
Read the worked answer
Available HCl = 0.0400 L x 2.00 mol/L = 0.0800 mol.
For the calculated composition, Mg requires 2 mol HCl per mol Mg: 0.01234 x 2 = 0.02468 mol. Al requires 3 mol HCl per mol Al: 0.007413 x 3 = 0.02224 mol. Total HCl required = 0.04692 mol, which is less than 0.0800 mol, so the acid is in excess.
If the acid were insufficient, the volume of hydrogen in the first experiment would be lower than the true value. The student would then calculate too low an apparent magnesium mass, possibly below 100%, and the result would no longer expose the aluminium interference. The NaOH experiment would still give the aluminium content, but the two experiments would not reconcile.