1. $n_{FeCl2}$ = $0,2.0,15=0,03$ ($mol)$
$FeCl_{2}$ + $2NaOH$ → $2NaCl$ + $Fe(OH)_{2}↓$
$0,03$ $0,06$ $0,06$ $0,03$ $(mol)$
$m=0,03.(56+(16+1).2)=2,7g$
$C_{M NaCl}$ = $\frac{0,06}{0,3}$ = $0,2M$
2. $n_{CuSO4}$ = $\frac{32.10%}{64+32+16.4}$= $0,02$ ($mol)$
$CuSO_{4}$ + $Zn$→ $Cu↓$ + $ZnSO_{4}$ ($phản$ $ứng$ $trao$ $đổi)$
$0,02$ $0,02$ $0,02$ $0,02$ $(mol)$
$m_{Zn}$ = $0,02.65=13g$