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Đáp án:

`Zn + 2HCl->ZnCl_{2} + H_{2}↑` (1)

`ZnO + 2HCl->ZnCl_{2} + H_{2}O` (2)

a) Ta có: `n_{H_{2}}=(V)/(22,4)=(3,36)/(22,4)= 0,15 (mol)`

Theo PT (1): `n_{Zn}=n_{H_{2}}= 0,15 (mol)`

`=> m_{Zn}=n × M=0,15 × 65= 9,75 (g)`

Mà: `m_{hh}=m_{Zn} + m_{ZnO}`

`=> m_{ZnO}=m_{hh} - m_{Zn}=17,85 - 9,75= 8,1 (g)`

b)

Theo PT (1): `n_{HCl}=2n_{Zn}=2 × 0,15= 0,3 (mol)`

Theo PT (2): `n_{HCl}=2n_{ZnO}=2 × (8,1)/(81)= 0,2 (mol)`

`=> m_{HCl}=(0,3 + 0,2) × 36,5= 18,25 (g)`

`=>` $m_{\text{dd HCl}}=\text{$\dfrac{m_{ct}}{\text{C%}}$ × 100%=$\dfrac{18,25}{12,5}$ × 100%= 146 (g)}$

c)

`m_{\text{dd sau pứ}}=17,85 + 146 - 0,15 × 2= 163,55 (g)`

Theo PT (1): `n_{ZnCl_{2}}=n_{Zn}= 0,15 (mol)`

Theo PT (2): `n_{ZnCl_{2}}=n_{ZnO}=(8,1)/(81)= 0,1 (mol)`

`=>` $\text{C%}_{ZnCl_{2}}=\text{$\dfrac{m_{ct}}{m_{dd}}$ × 100%=$\dfrac{(0,15 + 0,1) × 136}{163,55}$ × 100%= 20,8%}$