K + 1/2 Cl2 ---> KCl
1/2 moli Cl2.........74,5 g KCl
4 moli Cl2................x g KCl x = 596 g KCl
Fe + H2SO4 ---> FeSO4 + H2
pV=nr moli * R * T
=> nr moli = (12,3*2)/(300*0,082)=1 mol H2
1 mol Fe..........1 mol H2
x moli Fe...........1 mol H2 x = 1 mol Fe = 56 g Fe