2 Fe + 3 Cl2 => 2 FeCl3
m aliaj = 2.24 kg
%Fe = 25%
m Fe = m aliaj x 25/100 = 2.24 x 25/100 = 0.56 kg = 560 g Fe pur in aliaj
M Fe = 56 g/mol
n Fe = m Fe/ M Fe => n Fe = 560 / 56 = 10 mol Fe
2 mol Fe reactioneaza cu 3 mol Cl2
10 mol Fe reactioneaza cu a mol Cl2
a=10 x 3 / 2 = 15 mol Cl2