$\rm n_{Al}=\dfrac{5,4}{27}=0,2(mol)$ 

PTHH: $\rm 2Al +3H_2SO_4 \to Al_2(SO_4)_3 +3H_2$ 

Theo phương trình hóa học 

$\rm n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=0,3(mol)$ 

`a)` $\rm m_{H_2SO_4}=0,3.98=29,4(g)$ 

`->` $\rm C\%_{H_2SO_4}=\dfrac{29,4}{160}.100\%=18,375\%$ 

`b)` $\rm n_{H_2}=n_{H_2SO_4}=0,3(mol)$ 

`->` $\rm V_{H_2}=0,3.24,79=7,437(l)$ 

`c)` $\rm m_{ddspu}=5,4+160-0,3.2=164,8(g)$ 

$\rm n_{Al_2(SO_4)_3}=\dfrac{1]{2}n_{Al}=0,1(mol)$ 

`->` $\rm C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342}{164,8}.100\%=20,75\%$