`a)`

`m_{CR\ spu}=m_{Cu}=3(g)`

`->\%m_{Cu}=3/{21,6}.100\%=13,89\%`

`Hh:Zn:x(mol);Fe:y(mol)`

`->65x+56y=21,6-3=18,6(1)`

`Zn+H_2SO_4->ZnSO_4+H_2`

`Fe+H_2SO_4->FeSO_4+H_2`

`n_{H_2}=x+y={7,437}/{24,79}=0,3(2)`

`(1)(2)->x=0,2;y=0,1`

`m_{Zn}=0,2.65=13(g);m_{Fe}=0,1.56=5,6(g)`

`\%m_{Zn}={13}/{21,6}.100\%=60,19\%`

`\%m_{Fe}=100-60,19-13,89=25,92\%`

`b)`

`n_{H_2SO_4\ pu}=n_{H_2}=0,3(mol)`

`n_{ZnSO_4}=x=0,2(mol);n_{FeSO_4}=y=0,1(mol)`

`m_{dd\ sau}=18,6+175-0,3.2=193(g)`

`C\%_{H_2SO_4\ du}={175.20\%-0,3.98}/{193}.100\%=2,9\%`

`C\%_{ZnSO_4}={0,2.161}/{193}.100\%=16,68\%`

`C\%_{FeSO_4}={0,1.152}/{193}.100\%=7,88\%`

$m_{rắn }=m_{Cu}=3(g)$ 

`\to %m_{Cu}=\dfrac{3}{21,6}×100%=13,89%`

Đặt $a(mol)=n_{Zn},b(mol)=n_{Fe}$ 

Theo đề ta có pt $65a+56b=18,6(1)$ 

$PTHH :Zn +H_2SO_4 \xrightarrow {} ZnSO_4 +H_2$

$Fe +H_2SO_4 \xrightarrow{} FeSO_4 +H_2$ 

Theo pt ta có 

$n_{Zn}+n_{Fe}=n_{H_2}=\dfrac{7,437}{24,79}=0,3(mol)$ 

$⇔a+b=0,3(mol)(2)$ 

từ (1) và (2) suy ra 

$a=0,2(mol), b=0,1(mol)$ 

$\to m_{Fe}=5,6(g)$ 

`⇒%m_{Fe}=\dfrac{5,6}{21,6}×100% =25,92%`

`\to %m_{Zn}=100%-13,89%-25,92%=60,19%`

$b)$ 

$n_{H_2SO_4(pứ)}=n_{H_2}=0,3(mol)$ 

Theo pt 

$n_{ZnSO_4}=n_{Zn}=0,2(mol)$ 

$n_{FeSO_4}=n_{Fe}=0,1(mol)$ 

$m_{dds pứ}=18,6+175-0,3×2=193(g)$ 

`C% _{H_2SO_4(dư)}= \frac{175×20% -0,3×98}{193}×100%=2,9%`

`C%_{ZnSO_4}=\frac{0,2×161}{193}×100%=16,68%`

`C%_{FeSO_4}=\frac{0,1×152}{193}×100%=7,88%`