\(\begin{array}{l}
a)\\
Dat\,n_{Al_2O_3}=x(mol);n_{MgO}=y(mol);n_{CuO}=z(mol)\\
\to 102x+40y+80z=18,72(1)\\
500ml=0,5l\to n_{HCl}=0,5.1,5=0,75(mol)\\
CuO+H_2\xrightarrow{t^o}Cu+H_2O\\
Theo\,PT:\,n_{Cu}=n_{CuO}=z(mol)\\
\to 102x+40y+64z=16,8(2)\\
Al_2O_3+6HCl\to 2AlCl_3+3H_2O\\
MgO+2HCl\to MgCl_2+H_2O\\
CuO+2HCl\to CuCl_2+H_2O\\
Theo\,PT:\,6x+2y+2z=n_{HCl}=0,75(3)\\
(1)(2)(3)\to \begin{cases} x=0,06\\ y=0,075\\ z=0,12 \end{cases}\\
\to \begin{cases} \%m_{Al_2O_3}=\frac{0,06.102}{18,72}.100\%≈32,69\%\\ \%m_{MgO}=\frac{0,075.40}{18,72}.100\%≈16\%\\ \%m_{CuO}=100-16-32,69=51,31\% \end{cases}\\
b)\\
Theo\,PT:\,n_{AlCl_3}=0,12;n_{MgCl_2}=0,075;n_{CuCl_2}=0,12\\
\to\begin{cases} C_{M\,AlCl_3}=\frac{0,12}{0,5}=0,24M\\ C_{M\,MgCl_2}=\frac{0,075}{0,5}=0,15M\\ C_{M\,CuCl_2}=\frac{0,12}{0,5}=0,24M\end{cases}
\end{array}\)

$\#duaconngoaihanhlang$

$a,$

$PTHH$ $:$

$CuO$ $+$ $H_2$ $\xrightarrow{t^o}$ $Cu$ $+$ $H_2O$ $(1)$

$m_{\text{O (oxit phản ứng)}}$ $=$ $m_X$ $-$ $m_{\text{chất rắn sau phản ứng}}$ $=$ $1,92$ $(g)$

$n_{\text{O (oxit phản ứng)}}$ $=$ $\frac{1,92}{16}$ $=$ $0,12$ $(mol)$

$Gọi$ $n_{MgO}$ $=$ $x$ $(mol)$

        $n_{Al_2O_3}$ $=$ $y$ $(mol)$

        $n_{CuO}$ $=$ $z$ $(mol)$

$⇒$ $40x$ $+$ $102y$ $+$ $80y$ $=$ $18,72$ $(I)$

$Có$ $n_{\text{O (oxit phản ứng)}}$ $=$ $n_{CuO}$ $=$ $z$ $(mol)$

$⇒$ $z$ $=$ $0,12$ $(II)$

$PTHH$ $:$

$MgO$ $+$ $2HCl$ $→$ $MgCl_2$ $+$ $H_2↑$ $(2)$

$Al_2O_3$ $+$ $6HCl$ $→$ $2AlCl_3$ $+$ $3H_2↑$ $(3)$

$CuO$ $+$ $2HCl$ $→$ $CuCl_2$ $+$ $H_2↑$ $(4)$

$n_{HCl}$ $=$ $0,5$ $×$ $1,5$ $=$ $0,75$ $(mol)$

$Theo$ $(1)$ $,$ $(2)$ $,$ $(3)$ $:$ $n_{HCl}$ $=$ $2n_{MgO}$ $+$ $6n_{Al_2O_3}$ $+$ $2n_{CuO}$ $=$ $2x$ $+$ $6y$ $+$ $2z$ $(mol)$

$⇒$ $2x$ $+$ $6y$ $+$ $2z$ $=$ $0,75$ $(III)$

$Từ$ $(I)$ $,$ $(II)$ $,$ $(III)$ $⇒$ $\begin{cases} x = 0,075\\y = 0,06\\z = 0,12 \end{cases}$

$⇒$ $\%_{MgO}$ $=$ $\frac{0,075×40}{18,72}$ $×$ $100\%$ $=$ $16,026\%$

$⇒$ $\%_{Al_2O_3}$ $=$ $\frac{0,06×102}{18,72}$ $×$ $100\%$ $=$ $32,692\%$

$⇒$ $\%_{CuO}$ $=$ $\frac{0,12×80}{18,72}$ $×$ $100\%$ $=$ $51,282\%$

$b,$

$Theo$ $(2)$ $:$ $n_{MgCl_2}$ $=$ $n_{MgO}$ $=$ $0,075$ $(mol)$

$⇒$ $C_{M_{MgCl_2}}$ $=$ $\frac{0,075}{0,5}$ $=$ $0,15$ $M$

$Theo$ $(3)$ $:$ $n_{AlCl_3}$ $=$ $2n_{Al_2O_3}$ $=$ $0,12$ $(mol)$

$⇒$ $C_{M_{AlCl_3}}$ $=$ $\frac{0,12}{0,5}$ $=$ $0,24$ $M$

$Theo$ $(3)$ $:$ $n_{CuCl_2}$ $=$ $n_{CuO}$ $=$ $0,12$ $(mol)$

$⇒$ $C_{M_{CuCl_2}}$ $=$ $\frac{0,12}{0,5}$ $=$ $0,24$ $M$