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Đáp án:

 62 C

63 A

Giải thích các bước giải:

 \(\begin{array}{l}
62)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
hh:Zn(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,2\\
65a + 56b = 12,1
\end{array} \right.\\
 \Rightarrow a = b = 0,1\,mol\\
\% {m_{Fe}} = \dfrac{{0,1 \times 56}}{{12,1}} \times 100\%  = 46,28\% \\
63)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
m = {m_{hh}} - {m_{{H_2}}} \Rightarrow {m_{{H_2}}} = 7,8 - 7 = 0,8\,g\\
{n_{{H_2}}} = \dfrac{{0,8}}{2} = 0,4\,mol\\
hh:Mg(a\,mol),Al(b\,mol)\\
\left\{ \begin{array}{l}
24a + 27b = 7,8\\
a + 1,5b = 0,4
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,2\\
\% {m_{Al}} = \dfrac{{0,2 \times 27}}{{7,8}} \times 100\%  = 69,23\% 
\end{array}\)