$\rm n_{H_2SO_4}=\dfrac{ 49.20\%}{98}= 0,1(mol)$ 

`a)`Phương trình hóa học 

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

`b)` 

Theo phương trình hóa học 

$\rm n_{H_2}=n_{H_2SO_4}=0,1(mol)$ 

$\rm n_{Al}=\dfrac{2}{3}n_{H_2SO_4}=\dfrac{1}{15}(mol)$ 

`@`$\rm m_{Al}=\dfrac{1}{15}.27=1,8(g)$ 

`@`$\rm V_{H_2(đkc)}=0,1.24,79=2,479(l)$ 

`c)` 
$\rm m_{dung~dịch~sau~phản ~ứng}=49+1,8-0,1.2=50,6(g)$ 

$\rm n_{Al_2(SO_4)_3}=\dfrac{1}{3}n_{Al}=\dfrac{1}{30}(mol)$ 

`->`$\rm m_{Al_2(SO_4)_3}=\dfrac{1}{30}.342=11,4(g)$ 

`=>`$\rm C\%_{Al_2(SO_4)_3}=\dfrac{11,4}{50,6}.100\%=22,53\%$

`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`.

`m_(ctH_2SO_4)=20%*49=9,8g`.

`n_(H_2SO_4)=(9,8)/(1*2+32+16*4)=0,1mol`.

`n_(Al)=0,1*2:3=1/15mol`.

`m_(Al)=1/15*27=1,8g`.

`n_(H_2)=n_(H_2SO_4)=0,1mol`.

`V_(H_2)=0,1*24,79=2,479l(đkc)`.

`V_(H_2)=0,1*22,4=2,224(đktc)`.

`m_(ddspu)=49+1,8-0,1*2=50,6g`.

`n_(Al_2(SO_4)_3)=0,1:3=1/30mol`.

`m_(Al_2(SO_4)_3)=1/30*342=11,4g`.

`C%_(H_2SO_4)=(11,4)/(50,6)*100%~~22,53%`.