Em tham khảo nha:
\(\begin{array}{l}
1)\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2\,mol\\
{n_{HCl}} = \dfrac{{54,57 \times 20\% }}{{36,5}} \approx 0,3\,mol\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = {n_{Fe}} \text{ phản ứng } = {n_{FeC{l_2}}} = \dfrac{{0,3}}{2} = 0,15\,mol\\
{m_{{\rm{dd}}spu}} = 0,15 \times 56 + 54,57 - 0,15 \times 2 = 62,67g\\
{C_\% }FeC{l_2} = \dfrac{{0,15 \times 127}}{{62,67}} \times 100\% = 30,4\% \\
2)\\
2KMn{O_4} \xrightarrow{t^0} {K_2}Mn{O_4} + Mn{O_2} + {O_2}\\
3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
{n_{Fe}} = \dfrac{{8,4}}{{56}} = 0,15\,mol\\
{n_{{O_2}}} = 0,15 \times \frac{2}{3} = 0,1\,mol\\
{n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,2\,mol\\
{m_{KMn{O_4}}} = 0,2 \times 158 = 31,6g
\end{array}\)