Răspuns:
2Fe +3Cl2 =2FeCl3
a)2x56gFe......................2x162,5gFeCl3
20g Fe.............................X
X=20x2x162,5/2x56 =58gFeCl3
b) 2x56gFe........................3x22,4lCl2
20gFe...............................Y
Y=20x3x22,4/2x56=12l Cl2
c)n=58/56=1,03 moli
Explicație:
A Fe=56
MFeCl3=162,5
1mol de clor are volum de 22,4l
n=m/M