Em tham khảo nha :
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,944}}{{22,4}} = 0,31mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
27a + 56b = 10,52\\
\frac{3}{2}a + b = 0,31
\end{array} \right.\\
\Rightarrow a = 0,12;b = 0,13\\
{m_{Al}} = 0,12 \times 27 = 3,24g\\
\% Al = \dfrac{{3,24}}{{10,52}} \times 100\% = 30,8\% \\
\% Fe = 100 - 30,8 = 69,2\% \\
b)\\
C1:\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,12mol\\
{m_{A{l_2}{{(S{O_4})}_3}}} = 0,12 \times 133,5 = 16,02g\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,13mol\\
{m_{FeC{l_2}}} = 0,13 \times 127 = 16,51g\\
{m_m} = 16,02 + 16,51 = 32,53g\\
C2:\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,62mol\\
{m_{HCl}} = 0,62 \times 36,5 = 22,63g\\
\text{Theo định luật bảo toàn khối lượng ta có :}\\
{m_{hh}} + {m_{HCl}} = {m_m} + {m_{{H_2}}}\\
\Rightarrow {m_m} = 10,52 + 22,63 - 0,31 \times 2 = 32,53g
\end{array}\)
