Đáp án:

Bạn tham khảo lời giải ở dưới nhé ! 

Giải thích các bước giải:

Bài 1:

\(\begin{array}{l}
MgO + {H_2} \to Mg + {H_2}O\\
CuO + {H_2} \to Cu + {H_2}O\\
Mg + 2HCl \to MgC{l_2} + {H_2}
\end{array}\)

\(\begin{array}{l}
a)\\
{m_{Cu}} = 9,6g\\
 \to {n_{Cu}} = 0,15mol\\
 \to {n_{CuO}} = {n_{Cu}} = 0,15mol\\
 \to {m_{CuO}} = 12g\\
 \to {m_{MgO}} = 4g \to {n_{MgO}} = 0,1mol\\
 \to {n_{Mg}} = {n_{MgO}} = 0,1mol\\
 \to {m_X} = {m_{Mg}} + {m_{Cu}} = 12g
\end{array}\)

\(\begin{array}{l}
b)\\
{n_{{H_2}}} = {n_{MgO}} + {n_{CuO}} = 0,25mol\\
 \to {V_{{H_2}}} = 5,6l\\
c)\\
{n_{{H_2}O}} = {n_{MgO}} + {n_{CuO}} = 0,25mol\\
 \to {m_{{H_2}O}} = 4,5g
\end{array}\)

Bài 2: 

\(\begin{array}{l}
A{l_2}{O_3} + 3{H_2} \to 2Al + 3{H_2}O\\
CuO + {H_2} \to Cu + {H_2}O\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}
\end{array}\)

\(\begin{array}{l}
a/\\
{n_{HCl}} = 0,9mol\\
 \to {n_{Al}} = \frac{1}{3}{n_{HCl}} = 0,3mol\\
 \to {m_{Al}} = 8,1g\\
 \to {n_{A{l_2}{O_3}}} = \frac{1}{2}{n_{Al}} = 0,15mol\\
 \to {m_{A{l_2}{O_3}}} = 15,3g\\
 \to {m_{CuO}} = 8g \to {n_{CuO}} = 0,1mol\\
 \to {n_{Cu}} = {n_{CuO}} = 0,1mol\\
 \to {m_{Cu}} = 6,4g\\
 \to {m_X} = {m_{Al}} + {m_{Cu}} = 14,5g
\end{array}\)

\(\begin{array}{l}
b/\\
{n_{{H_2}}} = 3{n_{A{l_2}{O_3}}} + {n_{CuO}} = 0,55mol\\
 \to {V_{{H_2}}} = 12,32l
\end{array}\)