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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?

Respuesta :

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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%

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