Đáp án:
a)
$FeSO_4$ dư và Zn tác dụng hết
b)
\(\begin{array}{l}
C{\% _{FeS{O_4}d}} = 22,7\% \\
C{\% _{ZnS{O_4}}} = 8,01\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{Zn}} = \dfrac{m}{M} = \dfrac{{13}}{{65}} = 0,2mol\\
{m_{FeS{O_4}}} = \dfrac{{m \times 30,4\% }}{{100\% }} = \dfrac{{400 \times 30,4\% }}{{100\% }} = 121,6g\\
{n_{FeS{O_4}}} = \dfrac{m}{M} = \dfrac{{121,6}}{{152}} = 0,8mol\\
Zn + FeS{O_4} \to ZnS{O_4} + Fe\\
\dfrac{{0,2}}{1} < \dfrac{{0,8}}{1} \Rightarrow FeS{O_4}\text{ dư}\\
b)\\
{n_{FeS{O_4}d}} = 0,8 - 0,2 = 0,6mol\\
{m_{FeS{O_4}_d}} = n \times M = 0,6 \times 152 = 91,2g\\
{n_{ZnS{O_4}}} = {n_{Fe}} = {n_{Zn}} = 0,2mol\\
{m_{ZnS{O_4}}} = n \times M = 0,2 \times 161 = 32,2g\\
{m_{Fe}} = n \times M = 0,2 \times 56 = 11,2g\\
{m_{{\rm{dd}}}} = {m_{FeS{O_4}}} + {m_{Zn}} - {m_{Fe}} = 400 + 13 - 11,2 = 401,8g\\
\Rightarrow C{\% _{FeS{O_4}d}} = \dfrac{{{m_{FeS{O_4}d}}}}{{{m_{{\rm{dd}}}}}} \times 100\% = \dfrac{{91,2}}{{401,8}} \times 100\% = 22,7\% \\
C{\% _{ZnS{O_4}}} = \dfrac{{{m_{ZnS{O_4}}}}}{{{m_{{\rm{dd}}}}}} \times 100\% = \dfrac{{32,2}}{{401,8}} \times 100\% = 8,01\%
\end{array}\)
