Đáp án:
$Mg +1/2O_{2} → MgO$
$4Al +3O_{2} →2Al_{2}O_{3}$
$MgO+ H_{2}SO_{4} → MgSO_{4} + H_{2}O$
$Al_{2}O_{3} +3H_{2}SO_{4} → Al_{2}(SO_{4})_{3} +3H_{2}O$
$nMg=\frac{7,8.30,77\%}{24}=0,1$
$nMg=nMgO=nMgSO_{4}=0,1$
$nAl=\frac{7,8.(100-30,77)\%}{27}=0,2$
$nAl=2nAl_{2}(SO_{4})_{3}⇒nAl_{2}(SO_{4})_{3}=0,1$
$nAl_{2}(SO_{4})_{3}=nAl_{2}O_{3}=0,1$
$nH_{2}SO_{4}=\frac{400.14,6\%}{98}=0,6$
$nH_{2}SO_{4}phản ứng=nMgSO_{4}+3nAl_{2}(SO_{4})_{3}=0,4$
$nH_{2}SO_{4}dư=0,6-0,4=0,2$
$mdd=mH_{2}SO_{4} +mAl_{2}O_{3}+mMgO$
$mdd=400+0,1.102+0,1.40=414,2g$
$C\%MgSO_{4}=\frac{0,1.120}{414,2}.100=2,9\%$
$C\%Al_{2}(SO_{4})_{3}=\frac{0,1.342}{414,2}.100=8,26\%$
$C\%H_{2}SO_{4}dư=\frac{0,2.98}{414,2}.100=4,73\%$