mng giúp e vs ạ e cảm ơn nhiều ạ

$\rm n_{CuSO_4}=\dfrac{m_{dd}.C\%}{100.M}=\dfrac{20.10}{100.160}=0,0125(mol)$

$\rm PTHH:Zn+CuSO_4 \to ZnSO_4+Cu$
$\rm n_{Zn}=n_{ZnSO_4}=n_{Cu}=n_{CuSO_4}=0,0125(mol)$

$\rm m_{Zn}=0,0125.65=0,8125(g)$

$\rm m_{dd sau}=0,8125+20-0,0125.64=20,0125(g)$

$\rm C\% ZnSO_4=\dfrac{0,0125.161.100\%}{20,0125}=10,06\%$

`(m_(ct))/(m_(dd))*100%=C%`.

`=>10%=(m_(ct))/20*100%`.

`=>m_(ct)=2gam`.

`Zn+CuSO_4->ZnSO_4+Cu`.

`n_(CuSO_4)=(m_(CuSO_4))/(M_CuSO_4)=2/(64+32+16*4)=1/80mol=n_(Zn)=n_(ZnSO_4)=n_(Cu)`

`=>m_(Zn)=n_(Zn)*M_(Zn)=1/80*65=0,8125gam`.

`m_(dd sau pứ)=m_(Zn)-m_(CuSO_4)-m_(Cu)`.

`=>m_(dd sau pứ)=65*1/80+20-64*1/80`.

`=>m_(dd sau pứ)=20,0125gam`.

`m_(ct)=1/80*(65+32+16*4)=2,0125gam`.

`%C=(2,0125)/(20,0125)*100%~~10,06%`.