Em tham khảo nha:
\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{15,68}}{{22,4}} = 0,7\,mol\\
hh:Zn(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
65a + 56b = 43,7\\
a + b = 0,7
\end{array} \right.\\
\Rightarrow a = 0,5;b = 0,2\\
{m_{Zn}} = 0,5 \times 65 = 32,5g\\
{m_{Fe}} = 43,7 - 32,5 = 11,2g\\
b)\\
F{e_3}{O_4} + 4{H_2} \to 3Fe + 4{H_2}O\\
{n_{Fe}} = 0,7 \times \frac{3}{4} = 0,525\,mol\\
{m_{Fe}} = 0,525 \times 56 = 29,4g\\
c)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}(1)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}(2)\\
{n_{Al}} = \dfrac{a}{{27}}\,mol \Rightarrow {n_{{H_2}(1)}} = \dfrac{a}{{27}} \times \dfrac{3}{2} = \dfrac{a}{{18}}\,mol\\
{n_{Zn}} = \dfrac{b}{{65}}\,mol \Rightarrow {n_{{H_2}(2)}} = \dfrac{b}{{65}}\,mol\\
\Rightarrow \dfrac{a}{{18}} = \dfrac{b}{{65}} \Rightarrow a:b = 18:65
\end{array}\)