Răspuns:
a) 2Fe +3Cl2 => 2FeCl3 reacție de combinare
b)
MFe =56g/mol
MFeCl3= 56+3×35,5=162,5g/mol
m=n×M= 6×162,5=975g FeCl3
2×56g...............2×162,5g
2Fe +3Cl2 => 2FeCl3
xg.........................975g
x=975×2×56/2×162,5=calculează
calculam numarul de moli
2moli..................2moli
2Fe +3Cl2 => 2FeCl3
ymoli...............6moli
y=6×2/2=6 moli Fe
