`n_{Zn}=\frac{9,75}{65}=0,15(mol)`

`a)` `Zn+2HCl\to ZnCl_2+H_2`

`n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,15(mol)`

`n_{HCl}=2n_{Zn}=0,3(mol)`

`V_{H_2}=0,15.22,4=3,36(l)`

`b)` `C%_{HCl}=\frac{0,3.36,5.100%}{100}=10,95%`

`c)` Bảo toàn khối lượng:

`m_{ddspu}=m_{Zn}+m_{dd\ HCl}-m_{H_2}`

`=> m_{dd spu}=9,75+100-0,3=109,45g`

`C%_{ZnCl_2}=\frac{136.0,15.100%}{109,45}\approx 18,63%`

$n_{Zn}=\frac{9,75}{65} =0,15~(mol)$

$Zn+2HCl→ZnCl_2+H_2↑$

$0,15$     $0,3$          $0,15$       $0,15$         $(mol)$

$a.)$

$V_{H_2}=0,15.22,4=3,36~(lít)$

$b.)$

$mct_{HCl}= 0,3.(35,5+1)=10,95~(g)$

$C\ \%_{HCl}=\frac{10,95.100\ \%}{100}=10,95\ \%$

$c.)$

$mct_{ZnCl_2}= 0,15.(65+71)=20,4~(g)$

$mdd_{ZnCl_2}=9,75+100-0,3=109,45~(g)$

$C\ \%_{ZnCl_2}=\frac{20,4.100\ \%}{109,45}≈18,64\ \%$

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