se calculeaza Fe pur
p = mp.100 : mt p=80% [100-20=80 ]
mp=p .mt :100
mp=80.700: 100=560 g Fe PUR
560g yg xg
2 Fe +3 Cl2 =2FeCl3
2.56 g 3.71g 2.162,5g
x=560 . 2 .162,5: 2.56=1625 g FeCl3
y=560 .3.71 : 2.56=1065 g clor
se afla nr. de moli =n
n=m:M= 1065g : 71g/moli=15 moli clor
VCl2 =15moli . 22,4litri/moli=336 litri
MCl2=2.35,5=71------. 1mol=71g
MFeCl3=56+ 3.35,5=162,5 ------> 1mol=162,5g