$\rm n_{Al}=\dfrac{5,4}{27}=0,2(mol)$
$\rm n_{H_2SO_4}=\dfrac{9,8}{98}=0,1(mol)$
Phương trình hóa học
$\rm 2Al +3H_2SO_4 \to Al _2(SO_4)_3 +3H_2$
Tỉ lệ $\rm \dfrac{0,2}{2}>\dfrac{0,1}{3}$ `->` $\rm H_2SO_4$ hết
Theo phương trình hóa học
$\rm n_{H_2}=n_{H_2SO_4}=0,1(mol)$
`->` $\rm V_{H_2}=0,1.24,79=2,479(l)$
$\rm n_{Al(dư)}=0,2-\dfrac{2}{3}.0,1=\dfrac{2}{15}(mol)$
`->` $\rm m_{Al}=\dfrac{2}{15}.27=3,6(g)$
$\rm n_{Al_2(SO_4)_3}=\dfrac{1}{3}n_{H_2SO_4}=\dfrac{1}{30}(mol)$
`->` $\rm m_{Al_2(SO_4)_3}=\dfrac{1}{30}.342=11,4(g)$
$\rm m_{H_2}=0,1.2=0,2(g)$