`n_(H_2)=\frac{5,6}{22,4}=0,25(mol)`

`Cu` không tan trong `HCl`

`Zn+2HCl->ZnCl_2+H_2`

`0,25`     `0,5`                          `0,25`

`=>m_(Zn)=0,25.65=16,25(g)`

`m_(Cu)=29,05-16,25=12,8(g)`

`=>n_(Cu)=\frac{12,8}{64}=0,2(mol)`

`a,`

`%m_(Zn)=\frac{16,25}{29,05}.100=55,938%`

`%m_(Cu)=100-55,938=44,062%`

`b,`

`Zn+2H_2SO_4->ZnSO_4+SO_2+2H_2O`

`0,25`                                         `0,25`

`Cu+2H_2SO_4->CuSO_4+SO_2+2H_2O`

`0,2`                                             `0,2`

`V_(SO_2)=0,45.22,4=10,08(l)`

`c,`

`C_(MHCl)=\frac{0,5}{0,2}=2,5M`

a,

$n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)$

$Zn+2HCl\to ZnCl_2+H_2$

$\Rightarrow n_{Zn}=0,25(mol)$

$m_{Zn}=0,25.65=16,25g$

$\Rightarrow \%m_{Zn}=\dfrac{16,25.100}{29,05}=55,94\%$

$\%m_{Cu}=44,06\%$

b,

$n_{Cu}=\dfrac{29,05-16,25}{64}=0,2(mol)$

$Cu+2H_2SO_4\to CuSO_4+SO_2+2H_2O$

$\Rightarrow n_{SO_2}=0,2(mol)$

$\Rightarrow V=0,2.22,4=4,48l$

c,

$n_{HCl}=2n_{H_2}=0,5(mol)$

$\Rightarrow C_{M_{HCl}}=\dfrac{0,5}{0,2}=2,5M$