$n_{H_2SO_4}=0,2.3=0,6(mol)$ 

$n_{Al_2O_3}=\dfrac{120}{102}=1,18(mol)$ 

$\rm pthh : Al_2O_3 +3H_2SO_4 \xrightarrow{} Al_2(SO_4)_3 +3H_2O$ 

tỉ lệ $\dfrac{1,18}{1}>\dfrac{0,6}{3} $ `=>H_2SO_4` hết 

theo pthh 

$n_{Al_2(SO_4)_3} =\dfrac{1}{3}n_{H_2SO_4}=0,2(mol)$ 

`->`$m_{Al_2(SO_4)_3}=0,2. 342=68,4(g)$ 

$\rm n_{Al_2O_3}=\dfrac{120}{102}=1,18(mol)$

$\rm n_{H_2SO_4}=0,2.3=0,6(mol)$

$\rm a.PTHH:Al_2O_3+3H_2SO_4 \to Al_2(SO_4)_3+3H_2O$

ta có: $\rm {1,18}{1}> \dfrac{0,6}{3} \to Al_2O_3 dư$

$\rm n_{Al_2(SO_4)_3}=\dfrac{1}{3}n_{H_2SO_4}=0,2(mol)$

$\rm m_{Al_2(SO_4)_3}=0,2.342=68,4(g)$