Đáp án:

 `↓`

Giải thích các bước giải:

 $CuCl_{2}$ + `2NaOH` `→` $Cu(OH)_{2}$ + `2NaCl` 

- $n_{NaOH}$ `=` $\frac{m}{M}$ `=` $\frac{20}{40}$ `=` `0,5` `(mol)`

- Chất tham gia : $CuCl_{2}$ 

- Chất sản phẩm : $Cu(OH)_{2}$, `NaCl`

- Theo pt : $n_{CuCl2}$ `=` $\frac{1}{2}$ `×` `0,5` `=` `0,25` `(mol)`

`⇒` $m_{CuCl2}$ `=` `0,25` `×`  `135` `=` `33,75` `(g)`

- Theo pt : $n_{Cu(OH)2}$ `=` $\frac{1}{2}$ `×` `0,5` `=` `0,25` `(mol)`

`⇒` $m_{Cu(OH)2}$ `=` `0,25` `×` `98` `=` `24,5` `(g)`

- Theo pt : $n_{NaCl}$ `=` $n_{NaOH}$ `=` `0,5` `(mol)`

`⇒` $m_{NaCl}$ `=` `0,5` `×` `58,5` `=` `29,25` `(g)`

$n_{NaOH}=\dfrac{20}{40}=0,5(mol)\\ CuCl_2+2NaOH\to Cu(OH)_2+2NaCl\\ n_{CuCl_2}=n_{Cu(OH)_2}=\dfrac{1}{2}.n_{NaOH}=0,25(mol)\\ n_{NaCl}=n_{NaOH}=0,5(mol)\\ m_{CuCl_2}=0,25.135=33,75(g)\\ m_{NaCl}=0,5.58,5=29,25(g)\\ m_{Cu(OH)_2}=0,25.98=24,5(g)\\$