A 2.07-g sample of an iron-aluminum alloy (ferroaluminum) is dissolved in excess HCl (aq) to produce 0.100 g H2 (g).

Fe (s) + 2HCl (aq) →FeCl2 (aq) + H2 (g)

2Al (s) + 6HCl (aq) →2AlCl3 (aq) + 3H2 (g).

What is the percent composition, by mass, of the ferroaluminum?

Fe + 2 HCl - - > FeCl2 + H2

2 Al + 6 HCl - - > 2 AlCl3 + 3 H2

total moles H2 = 0.100 g / (2.016 g/mol) = 0.0496 mol

let x = mass Fe

let y = mass Al

Total mass of alloy is:

x + y = 2.07

Performing H2 balance based on stoichiometry:

x / 55.847 + 3 y/2 / 26.9815 = 0.0496

x / 55.847 + 3y/53.963 = 0.0496

53.963 x + 167.541 y = 149.48

53.967 (2.07 - y) + 167.541 y = 149.48

111.71 - 53.967 y + 167.541 y = 149.48

113.6 y = 37.77

y = 0.332 g = mass Al

% Al = 0.332 x 100 / 2.07 = 16.06 %

% Fe = 83.94%