Đáp án:

 a) 13g và 8,1 g

b) 40,8g

c) 75g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_{Zn}} = {n_{{H_2}}} = 0,2\,mol\\
{m_{Zn}} = 0,2 \times 65 = 13g\\
{m_{ZnO}} = 21,1 - 13 = 8,1g\\
b)\\
{n_{ZnO}} = \dfrac{{8,1}}{{81}} = 0,1\,mol\\
{n_{ZnC{l_2}}} = 0,2 + 0,1 = 0,3\,mol\\
{m_{ZnC{l_2}}} = 0,3 \times 136 = 40,8g\\
c)\\
{n_{HCl}} = 0,2 \times 2 + 0,1 \times 2 = 0,6\,mol\\
{m_{{\rm{dd}}HCl}} = \dfrac{{0,6 \times 36,5}}{{14,6\% }} = 150g
\end{array}\)

`n_(H_2)=\frac{4,48}{22,4}=0,2(mol)`

 `Zn+2HCl->ZnCl_2+H_2`

`ZnO+2HCl->ZnCl_2+H_2O`

Theo PT

`n_(Zn)=n_(O_2)=0,2(mol)`

`=>m_(Zn)=0,2.65=13 

`⇒m_(ZnO)=8,1(g)`

`=>n_(ZnO)=\frac{8,1}{81}=0,1(mol)`

`b,`

Theo PT

`n_(ZnCl_2)=n_(Zn)+n_(ZnO)=0,3(mol)`

`=>m_(ZnCl_2)=0,3.136=40,8(g)`

`c,`

Theo PT

`n_(HCl)=2(n_(Zn)+n_(ZnO))=0,6(mol)`

`=>m_(dd  HCl)=\frac{0,6.36,5}{14,6%}=150(g)`