Em tham khảo nha :

\(\begin{array}{l}
CuS{O_4} + Zn \to ZnS{O_4} + Cu\\
{n_{CuS{O_4}}} = \dfrac{{250 \times 20\% }}{{160}} = 0,3125mol\\
{n_{Zn}} = \dfrac{{6,5}}{{65}} = 0,1mol\\
\dfrac{{0,3125}}{1} > \dfrac{{0,1}}{1} \Rightarrow CuS{O_4}\text{ dư}\\
{n_{Cu}} = {n_{Zn}} = 0,1mol\\
{n_{ZnS{O_4}}} = {n_{Zn}} = 0,1mol\\
{m_{ZnS{O_4}}} = 0,1 \times 161 = 16,1g\\
{n_{CuS{O_4}d}} = 0,3125 - 0,1 = 0,2125mol\\
{m_{CuS{O_4}d}} = 0,2125 \times 160 = 34g\\
C{\% _{ZnS{O_4}}} = \dfrac{{16,1}}{{250 + 6,5 - 6,4}} \times 100\%  = 6,44\% \\
C{\% _{CuS{O_4}}} = \dfrac{{34}}{{250 + 6,5 - 6,4}} \times 100\%  = 13,6\% 
\end{array}\)

Đáp án+Giải thích các bước giải:

$n_{CuSO_4}=\dfrac{250.20\%}{160}=0,3125(mol)\\ n_{Zn}=\dfrac{6,5}{65}=0,1(mol)\\ Zn+CuSO_4\to ZnSO_4+Cu\\ n_{Zn}<n_{CuSO_4}\\ \to Zn\ hết\\ n_{ZnSO_4}=n_{Zn}=n_{Cu}=0,1(mol)\\ n_{CuSO_4\ dư}=0,3125-0,1=0,2125(mol)\\ m_{\text{dd sau}}=250+6,5-0,1.64=250,1(g)\\ C\%_{dd\ ZnSO_4}=\dfrac{0,1.161}{250,1}.100\%=6,44\%\\ C\%_{dd\ CuSO_4\ dư}=\dfrac{0,2125.160}{250,1}.100\%=13,59\%$