Đặt $n_{MgO}$=a(mol),$n_{CuO}$=b(mol),$n_{Fe2O3}$=c(mol)

⇒$m_{X}$=40a+80b+160c=6(g) (I)

Có:$m_{O(X)}$=29,33% . 6=1,76(g)

⇒ a+b+3c=$\frac{1,76}{16}$ =0,11 

⇒ 18a+18b+54c=1,98 (II)

(I)(II)⇒58a+98b+214c=7,98

$\left \{ {{MgO,CuO} \atop {Fe2O3}} \right.$ + $H_{2}$$SO_{4}$ →$\left \{ {{MgSO4,CuSO4} \atop {Fe2(SO4)3}} \right.$ + $Ba(OH)_{2}$ →$\left \{ {{Mg(OH)2,Cu(OH)2} \atop {Fe(OH)3,BaSO4}} \right.$ 

Có: $m_{Mg(OH)2}$+ $m_{Cu(OH)2}$+$m_{Fe(OH)3}$=58a+98b+214c=7,98

       $n_{SO4}$= $n_{MgSO4}$+ $n_{CuSO4}$+$n_{Fe2(SO4)3}$=a+b+3c=0,11 mol

⇒$m_{BaSO4}$=0,11.233=25,63 g

⇒∑$m_{↓}$=25,63+7,98=33,61 (g)

$n_O=\dfrac{6.29,33\%}{16}=0,11(mol)$

$O+H_2SO_4\to SO_4+H_2O$

$\to n_{H_2SO_4}=n_{SO_4}=n_O=0,11(mol)$

$\to n_{BaSO_4\downarrow}=n_{SO_4}=0,11(mol)$

Gốc $SO_4$ hoá trị II còn gốc $OH$ (trong bazơ) hoá trị I nên ta có:

$2n_{OH}=n_{SO_4}\to n_{OH}=0,055(mol)$

Kết tủa gồm 3 bazơ và $BaSO_4$

$m_{KL}=6-0,11.16=4,24g$

$\to m=4,24+0,055.17+0,11.233=30,805g$