m(Al)=5.4г
Объяснение:
2Аl+3Cl2=2AlCl3
n(Cl2)=V/Vm=6.72÷22.4=0.3 моль
n(Al)=n(Cl2)×2/3=0.3×2/3=0.2моль
m(Al)=n×M=0.2×27=5.4г
