Fe + 2HCl -> FeCl2 + H2
b) MFeCl2=127g/mol
n=m/M=144/127=1,1338 moli FeCl2
1 mol FeCl2..................NA=6,023*10^23 molecule
1.1338 moli FeCl2..............x molecule
x=6,023*10^23*1,1338= Calculeaza !!
c) MFe=56g/mol
n=m/M=80/56=1,428 moli Fe
n Fe=nFeCl2 = 1,428 moli FeCl2
m FeCl2=1,428*127=181,428g FeCl2