This is a problem with two equations and two unknowns and you solve them simultaneously. But first you didn't carry out the calculation for mols H2; you threw away some. I solved for n and obtained 0.00719879 which I would round to 0.0072 mols H2.
Let X = mass Fe
and Y = mass Al
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equation 1 is X + Y = 0.1924
For equation 2 we need to make it so that mols H2 from Fe + mols H2 from Al = total mols = 0.0072. The equations are
Fe + 2HCl ==> FeCl2 + H2
2Al + 6HCl ==> 2AlCl3 + 3H2
To save typing space I'll let am stand for atomic mass.
mols H2 from Fe = X(1/am Fe)(1 mol H2/1 mol Fe) = X/55.85
mols H2 from Al = Y(1/26.98)*(3 mols H2/2 mols Al) = 3Y/2*26.98) and equation 2 becomes
(X/55.85) + (3Y/2*26.98) = 0.0072
Solve those two equation simultaneously for X and Y, then
%Fe = (mass Fe/mass sample)*100 = (X/0.1924)*100 = ?
Al and Fe react with HCl to produce a chloride salt and hydrogen gas. A .1924g sample of a mixture of Al and Fe is treated with excess HCl solution. A volume of 159mL h2 gas is collected over water at 19C and 841 torr. What is the mass percent by mass of Fe in the mixture? The vapor pressure of water is at 19C is 16.5 torr.
I calculated the moles of H2 to be .007mol. Where do I go from here?
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