Câu 3:
$Fe_{}+2HCl$ ---> $FeCl_{2}+H_{2}$
$Cu(OH)_{2}+2HCl$ ---> $CuCl_{2}+2H_{2}O$
$2HCl+Na_{2}CO_{3}$ ---> $2NaCl+H_{2}O+CO_{2}$
$2Fe+3Cu(OH)_{2}$ ---> $2Fe(OH)_{3}+3Cu$
Câu 4:
$FeCl_{2}+2NaOH$ ---> $Fe(OH)_{2}↓+2NaCl$
$n_{FeCl_{2}}=0,2×0,15=0,03mol$
⇒ $n_{Fe(OH)_{2}}=0,03mol$
⇒ $m=m_{Fe(OH)_{2}}=90×0,03=2,7g$