$Mưa$
a,$m_{Fe2(SO4)3}$ = $n.M$ =$0,6.(56.2+32.3+16.4.3)$ = $240g$
b,$m_{Al2(SO4)3}$ = $n.M$ = $0,7.(27.2+32.3+16.4.3)$ = $239,4g$
c,$m_{K3PO4}$ = $0,35.(39.3+31+16.4)$ = $74,2g$
d,$m_{Fe(OH)3}$ = $n.M$ = $0,15.(56+16.3+1.3)$ = $16,05g$
e,$n_{Cl2}$ =$\frac{V}{22,4}$ =$\frac{10,08}{22,4}$ =$0,45mol$
$m_{Cl2}$ = $n.M$= $0,45.(35,5.2)$ = $31,95g$
Xin lổi bn nhìu lắm a , mk chỉ bik lm tới đây thui huhu :<