x g 25,4g
Fe + 2HCl --> FeCl2 + H2
56 127
M.FeCl2 = 56+35,5x2 = 127 g/mol
=> x = 56x25,4/127 = 11,2 g Fe necesar consumarii HCl
daca 11,2g Fe ....................... 100%
m g ............................. (100+10%exces)
=> m = 11,2x110/100 = 12,32 g Fe introdus in reactie