Fe + 2 HCl => FeCl2 + H2
56 g Fe........2*36,5 g HCl
2 g Fe............x g HCl
x = 2.6 g HCl
c = md/ms * 100 => ms = md/c * 100 = 2,6/18.25 * 100 = 14.246%
56 g Fe...........22,4 l H2
2 g Fe.............x l H2
y = 0,8 l H2
56 g Fe..........127 g FeCl2
2 g Fe.............z g FeCl2
z = 4.535 g FeCl2