$a)$Gọi $\begin{cases} n_{Al_{2}O_{3}}=x(mol)\\n_{CuO}=y(mol)\\n_{MgO}=z(mol) \end{cases}$
Các $PTHH$ xảy ra:
$Al_{2}O_{3}+2NaOH→2NaAlO_{2}+H_{2}O$
$NaAlO_{2}+CO_{2}+2H_{2}O→NaHCO_{3}+Al(OH)_{3}↓$
$Al_{2}O_{3}+6HCl→2AlCl_{3}+3H_{2}O(I)$
$CuO+2HCl→CuCl_{2}+H_{2}O(II)$
$MgO+2HCl→MgCl_{2}+H_{2}O(III)$
$b)$ $m_{X}=102x+80y+40z=21,65(g)(1)$
BTNT $[Al]$: $n_{Al(OH)_{3}}=2n_{Al_{2}O_{3}}=2x(mol)$
⇔$2x.78=11,7$
⇔156x=11,7(2)$
Từ PT $(II)$ và $(III)$
⇒$n_{HCl}=2y+2z=0,5(mol)(3)$
Giải hệ $(1)$,$(2)$ và $(3)$⇒$x=0,075;y=0,1;z=0,15(mol)$
⇒Trong $21,65g$ $X$ có
$m_{Al_{2}O_{3}}=0,075.102=7,65(g)$
$m_{CuO}=0,1.80=8(g)$
$m_{MgO}=0,15.40=6(g)$
⇒Trong $4,33g$ $X$ có
$m_{Al_{2}O_{3}}=\frac{7,65.4,33}{21,65}=1,53(g)$
$m_{CuO}=\frac{8.4,33}{21,65}=1,6(g)$
$m_{MgO}=\frac{6.4,33}{21,65}=1,2(g)$
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