Đáp án:

 26 C

27 C

28 A

Giải thích các bước giải:

\(\begin{array}{l}
26)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
\text{ Gọi a là số mol $H_2$ }\\
{n_{HCl}} = 2{n_{{H_2}}} = 2a\,mol\\
BTKL:{m_{hh}} + {m_{HCl}} = {m_m} + {m_{{H_2}}}\\
 \Rightarrow 11,6 + 2a \times 36,5 = 36,45 + 2 \times a \Rightarrow a = 0,35\,mol\\
{V_{{H_2}}} = 0,35 \times 224 = 7,84l\\
27)\\
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}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
\text{ Gọi a là số mol $H_2$ }\\
{n_{{H_2}S{O_4}}} = {n_{{H_2}}} = a\,mol\\
BTKL:{m_{hh}} + {m_{{H_2}S{O_4}}} = {m_m} + {m_{{H_2}}}\\
 \Rightarrow 16,2 + a \times 98 = 69 + 2 \times a \Rightarrow a = 0,55\,mol\\
{V_{{H_2}}} = 0,55 \times 224 = 12,32l\\
28)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{8,288}}{{22,4}} = 0,37\,mol\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,74\,mol\\
BTKL:{m_{hh}} + {m_{HCl}} = {m_m} + {m_{{H_2}}}\\
 \Rightarrow m + 0,74 \times 36,5 = 44,87 + 2 \times 0,37 \Rightarrow m = 18,6g\\

\end{array}\)