Đáp án:

 a) 1,02mol

b) 51,93g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2Mg + {O_2} \to 2MgO\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
2Zn + {O_2} \to 2ZnO\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
ZnO + 2HCl \to ZnC{l_2} + {H_2}O\\
BTKL:\\
{m_{hh}} + {m_{{O_2}}} = {m_A} \Rightarrow {m_{{O_2}}} = 23,88 - 15,72 = 8,16g\\
hh:Mg(a\,mol),Al(b\,mol),Zn(c\,mol)\\
{n_{{O_2}}} = \dfrac{{8,16}}{{32}} = 0,255\,mol\\
 \Rightarrow 0,5a + 0,75b + 0,5c = 0,255\,mol\\
{n_{HCl}} = 2a + 3b + 2c \Rightarrow {n_{HCl}} = 4{n_{{O_2}}} = 1,02\,mol\\
b)\\
{n_{{H_2}O}} = \dfrac{{1,02}}{2} = 0,51\,mol\\
{m_m} = 23,88 + 1,02 \times 36,5 - 0,51 \times 18 = 51,93g
\end{array}\)