Đáp án:
$nHCl=\frac{3,65}{36,5}=0,1$
Gọi số mol $MgCl_{2}$x : mol ; $AlCl_{3}$: y mol
$40x+51y=1,82$
$2x+3y=0,1$
⇒$\left \{ {{x=0,02} \atop {y=0,02}} \right.$
$nMgO=nMgCl_{2}=0,02$
$mMgO=0,02.40=0,8g$
$mAl_{2}O_{3}=1,82-0,8=1,02g$
$C\%MgCl_{2}=\frac{0,02.95}{1,82+3,65}.100=34,73\%$
$C\%AlCl_{3}=\frac{0,02.133,5}{1,82+3,65}.100=48,81\%$
$nZn=\frac{1,3}{65}=0,02$
$2nZn=2nH_{2}⇒nH_{2}=0,02$
$VH_{2}=0,02.22,4=0,448lit$