`#Ly`

$a)$

`n_(H_2` `=` `(3,36)/(22,4)=0,15` $(mol)$

PTHH:

$Zn$ $+$ $2HCl$ `->` $ZnCl_2$ $+$ $H_2$
$ZnO$ $+$ $2HCl$ `->` $ZnCl_2$ $+$ $H_2O$

`->` `n_(Zn` `=` `n_(H_2` `=` `0,15` $(mol)$

`m_(Zn` `=` `0,15.65=9,75` $(g)$

`m_(ZnO` `=` `17,85-9,75=8,1` $(g)$

$b)$

Ta có: `n_(HCl` `=` `2n_(H_2` `=` `0,3` $(mol)$

`n_(ZnO` `=` `(8,1)/(81)=0,1` $(mol)$

`->` `n_(HCl` `=` `2n_(ZnO` `=` `0,2` $(mol)$

`∑``n_(HCl` `=` `0,3+0,2=0,5` $(mol)$

`m_(HCl` `=` `0,5.36,5=18,25` $(g)$

`m_(ddHCl` `=` `(18,25.100)/(12,5)=146` $(g)$

$c)$

Theo PT, `n_(ZnCl_2` `=` `(1)/(2)n_(HCl` `=` `0,25` $(mol)$

`m_(ZnCl_2` `=` `0,25.136=34` $(g)$

`m_(ddZnCl_2` `=` `17,85+146-0,15.2=163,55` $(g)$

`C%_(ZnCl_2` `=` `(34)/(163,55).100%≈20,79%`

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%}$