$n_{Zn}=\dfrac{13}{65}=0.2(mol);n_{Fe}=\dfrac{14}{56}=0.25(mol)$

$n_{H_2SO_4}=0.5 (mol)$

$m_{dd,H_2SO_4}=500 \times 1.14=570(gam)$

$a.$ 

$Zn+H_2SO_4 \to ZnSO_4+H_2$

$0.2$`->`$0.2$ `->`$0.2$  `->`$0.2$

$Fe+H_2SO_4 \to FeSO_4+H_2$

$0.25$`->`$0.25$ `->`$0.25$  `->`$0.25$

$b.$

$n_{H_2}=0.2+0.25=0.45(mol)$

`=>`$V_{H_2}=0.45 \times 22.4=10.08(L)$

$c.$

$m_{dd,sau}=13+14+570-0.45 \times 2=596.1(gam)$

`->`$m_{H_2O,thêm}=600-591.6=3.9(gam)$

$n_{H_2SO_4,dư}=0.5-0.45=0.05(mol)$

$C\%(H_2SO_4)=\dfrac{0.05 \times (2+96)}{600} \times 100\%=0.82\%$

$C\%(ZnSO_4)=\dfrac{0.2 \times (65+96)}{600} \times 100\%=5.37\%$

$C\%(FeSO_4)=\dfrac{0.25 \times (65+96)}{600} \times 100\%=6.33\%$

`a)`

`Zn+H_2SO_4->ZnSO_4+H_2`

`Fe+H_2SO_4->FeSO_4+H_2`

`b)`

`n_{Zn}={13}/{65}=0,2(mol)`

`n_{Fe}={14}/{56}=0,25(mol)`

`->n_{H_2}=n_{Fe}+n_{Zn}=0,45(mol)`

`V_{H_2}=0,45.22,4=10,08(L)`

`c)`

`n_{H_2SO_4\ pu}=n_{H_2}=0,45(mol)`

`n_{H_2SO_4\ du}=0,5-0,45=0,05(mol)`

`n_{ZnSO_4}=n_{Zn}=0,2(mol);n_{FeSO_4}=n_{Fe}=0,25(mol)`

`m_{dd\ sau}=500.1,14+13+14-0,45.2=596,1(g)`

`->m_{H_2O\ them}=600-596,1=3,9(g)`

`C\%_{ZnSO_4}={0,2.161}/{600}.100\%=5,37\%`

`C\%_{FeSO_4}={0,25.152}/{600}.100\%=6,33\%`

`C\%_{H_2SO_4\ du}={0,05.98}/{600}.100\%=0,82\%`