Amestec echimolecular => nr moli Fe = nr moli Al
Al + 3 HCl ---> AlCl3 + 3/2 H2
Fe + 2 HCl ---> FeCl2 + H2
1 mol Al...........3/2 moli H2
x moli Al..........a moli H2
1 mol Fe..........1 mol H2
x moli Fe...........b moli H2
a+b = 0,5 => 3/2x+x=0,5 => 5x=1=>x = 0,2 moli -> 0,2 moli Al si 0,2 moli Fe
masa amestec = 0,2*27+0,2*54 = 5,4+10,8=16,2 g
%Al = 5,4/16,2 * 100 = 33,333 %
%Mg = 10.8/16,2 * 100 = 66,666 %