Em tham khảo nha :
\(\begin{array}{l}
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
CaO + 2HCl \to CaC{l_2} + {H_2}O\\
{n_{HCl}} = 0,08 \times 1 = 0,08mol\\
hh:F{e_2}{O_3}(a\,mol),CaO(b\,mol)\\
\left\{ \begin{array}{l}
160a + 56b = 2,16\\
6a + 2b = 0,08
\end{array} \right.\\
\Rightarrow a = 0,01;b = 0,01\\
{m_{F{e_2}{O_3}}} = 0,01 \times 160 = 1,6g\\
\% F{e_2}{O_3} = \dfrac{{1,6}}{{2,16}} \times 100\% = 74,07\% \\
\% CaO = 100 - 74,07 = 25,93\% \\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,02mol\\
{m_{FeC{l_3}}} = 0,02 \times 162,5 = 3,25g\\
{n_{CaC{l_2}}} = {n_{CaO}} = 0,01mol\\
{m_{CaC{l_2}}} = 0,01 \times 111 = 1,11g\\
m = 3,25 + 1,11 = 4,36g
\end{array}\)