a,
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$
$MgO+H_2SO_4\to MgSO_4+H_2O$
$n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$
$\Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,2(mol)$
$\Rightarrow \%m_{Al}=\dfrac{0,2.27.100}{9,4}=57,45\%$
$\%m_{MgO}=42,55\%$
b,
$n_{Al_2(SO_4)_3}=0,1(mol)$
Ta có: $27n_{Al}+40n_{MgO}=9,4$
$\Rightarrow n_{MgO}=0,1(mol)$
$\Rightarrow n_{MgSO_4}=0,1(mol)$
$Al_2(SO_4)_3+4Ba(OH)_2\to 3BaSO_4+Ba(AlO_2)_2+4H_2O$
$MgSO_4+Ba(OH)_2\to BaSO_4+Mg(OH)_2$
Sau các phản ứng:
$n_{BaSO_4}=0,1.3+0,1=0,4(mol)$
$n_{Mg(OH)_2}=0,1(mol)$
$\to m=0,4.233+0,1.58=99g$