Em tham khảo nha:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Fe + 2HCl \to MgC{l_2} + {H_2}\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
{n_{Mg}} = {n_{Fe}} = {n_{F{e_2}{O_3}}} = a\,mol\\
\Rightarrow 24a + 56a + 160a = 2,4 \Rightarrow a = 0,01\,mol\\
{n_{HCl}} = 0,01 \times 2 + 0,01 \times 2 + 0,01 \times 6 = 0,1\,mol\\
{m_{HCl}} = n \times M = 0,1 \times 36,5 = 3,65g\\
{C_\% }(HCl) = \dfrac{{3,65}}{{500}} \times 100\% = 0,73\% \\
b)\\
{n_{{H_2}}} = 0,01 + 0,01 = 0,02\,mol\\
{m_{{\rm{dd}}spu}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 2,4 + 500 - 0,02 \times 2 = 502,36g\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,01\,mol\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,01\,mol\\
{n_{FeC{l_3}}} = 2{n_{F{e_2}{O_3}}} = 0,02\,mol\\
{C_\% }(MgC{l_2}) = \dfrac{{0,01 \times 95}}{{502,36}} \times 100\% = 0,19\% \\
{C_\% }(FeC{l_2}) = \dfrac{{0,01 \times 127}}{{502,36}} \times 100\% = 0,25\% \\
{C_\% }(FeC{l_3}) = \dfrac{{0,02 \times 162,5}}{{502,36}} \times 100\% = 0,65\%
\end{array}\)
