1. masa pura / masa totala = puritatea
masa pura Fe = 90/100 × 6 = 5.4g
2.
112........219...........................6
2Fe + 6HCl = 2 FeCl3 + 3H2
5.4..........x...............................y
3.
x = 5.4 × 219 /112 = 10.56 g
10.56 / masa sol = 5/100
masa sol = 211.2 g
4.
y = 6×5.4/112 = 0.29 g
nr moli H2 = 0.29 /2 = 0.145 moli
vol = 0.145 × 22.4 = 3.25 l