Răspuns :
Răspuns:
MFe=56g/mol
MFeCl3 =56+3×35,5=162,5g/mol
100 g Fe ...............................80 g Fe pur
700 g Fe .................................x g Fe pur
x = 700×80/100 = 560g Fe pur
2Fe + 3Cl2 => 2FeCl3
2×56 g Fe ................... 2×162,5 g FeCl3
560 g Fe........................... y g FeCl3
y = 560×2×162,5/2×56 = 1625 g FeCl3