Question

A co-op student designed an experiment to collect hydrogen gas by reacting iron with a hydrochloric acid (HCl) solution. The student’s co-op supervisor suggested that
aluminum can also be included to generate hydrogen gas:
Fe + 2 HCl → 2 FeCl2 + H2
2 Al + 6 HCl → 2 AlCl3 + 3 H2
The co-op student took a 0.4 g sample of a mixture of aluminum and iron and treated them with excess HCl solution. After completion of the reaction, a total volume of 305 mL of hydrogen dry gas was produced at 20oC and 833.5 mmHg. What is the percent by mass of iron in the original mixture?

Answer 1

This is much like the C3H8/C4H10 problem but perhaps a little easier to follow.
First I would conver the H2 gas in the PV = nRT formula and solve for n = number of mols. I obtained 0.00139 mols H2 gas at the conditions listed. I didn't know how far to carry the decimal; I assume you rounded the zeros on the 0.4g sample.
Let X = grams Fe
and Y = grams Al
--------------------
equation 1 is X + Y = 0.400 g

equation 2 is mols H2 produced by Fe + mols H2 produced by Al = 0.0139 mols.
mols H2 from Fe = (X/55.85)*(1 mol H2/1 mol Fe) = just X/55.85.
mols H2 from Al = (Y/26.98)*(3 mols H2/2 mols Al) = (3Y/2*26.98). Put that together for equation 2 of
(X/55.85) + (3Y/2*26.98) = 0.0139

Solve for X and Y simultaneously, then
%Fe = (X/0.400)*100 = ?