Trust gửi ạ

Em tham khảo nha:

\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}(1)\\
Fe + 2HCl \to FeC{l_2} + {H_2}(2)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}(3)\\
\text{ Gọi m là khối lượng của ba chất }\\
{n_{Mg}} = \dfrac{m}{{24}}mol \Rightarrow {n_{{H_2}(1)}} = {n_{Mg}} = \dfrac{m}{{24}}\,mol\\
{n_{Fe}} = \dfrac{m}{{56}}\,mol \Rightarrow {n_{{H_2}(2)}} = {n_{Fe}} = \dfrac{m}{{56}}\,mol\\
{n_{Al}} = \dfrac{m}{{27}}\,\,mol \Rightarrow {n_{{H_2}(3)}} = \dfrac{m}{{27}} \times \dfrac{3}{2} = \dfrac{m}{{18}}\,mol\\
 \Rightarrow \text{ Al cho nhiều $H_2$ nhất } \\
b)\\
Mg + 2HCl \to MgC{l_2} + {H_2}(1)\\
Fe + 2HCl \to FeC{l_2} + {H_2}(2)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}(3)\\
\text{ Gọi a là số mol $H_2$ }\\
{n_{Mg}} = {n_{{H_2}(1)}} = a\,mol \Rightarrow {m_{Mg}} = 24a(g)\\
{n_{Fe}} = {n_{{H_2}(2)}} = a\,mol \Rightarrow {m_{Fe}} = 56a(g)\\
{n_{Al}} = {n_{{H_2}(3)}} \times \dfrac{2}{3} = \dfrac{{2a}}{3}\,mol \Rightarrow {m_{Al}} = \dfrac{{2a}}{3} \times 27 = 18a(g)\\
\text{ Al tiết kiệm nhất }
\end{array}\)