Zn + 2HCl -> ZnCl2 + H2                  Fe + 2HCl -> FeCl2 + H2

 a                                  a                      b                                b

nZn = a mol và nFe = b mol

=> 65a + 56b = 43.7

a + b = 15.68/22.4 = 0.7

a = 0.5 và b = 0.2

mZn = 0.5 x 65 = 32.5g

mFe = 0.2 x 56 = 11.2g

4H2 + Fe3O4 -> 3Fe + 4H2O

 0.7      0.175      0.525

nFe3O4 = 0.2 mol

nH2 = 0.7 mol => Fe3O4 dư. H2 hết

nFe = 0.525 mol => mFe = 0.525 x 56 = 29.4g

2Al + 6HCl -> 2AlCl3 + 3H2                   Zn + 2HCl -> ZnCl2 + H2

 a/27                             3a/54                  b/65                             b/65

lượng H2 = nhau => 3a/54 = b/65

=> a/b = 18/65

Em tham khảo nha:

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