$m_{dd\ H_2SO_4}=560.1,4=784g$
$=>m_{H_2SO_4}=784.5\%=39,2g$
$=>n_{H_2SO_4}=\dfrac{39,2}{98}=0,4mol$
$n_{Al}=\dfrac{5,4}{27}=0,2mol$
$a.PTHH :$
$2Al + 3H_2SO_4\to Al_2(SO_4)_3+3H_2$
$\text{Theo pt : 2 mol 3 mol}$
$\text{Theo đbài : 0,2 mol 0,4 mol}$
Tỷ lệ : $\dfrac{0,2}{2}<\dfrac{0,4}{3}$
$\text{=>Sau pư H2SO4 dư}$
$\text{Theo pt :}$
$n_{H_2SO_4\ pư}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3mol$
$=>n_{H_2SO_4\ dư}=0,4-0,3=0,1mol$
$=>m_{H_2SO_4\ dư}=0,1.98=9,8g$
$\text{b,Theo pt :}$
$n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3mol$
$=>V_{H_2}=0,3.22,4=6,72l$
$\text{c.Theo pt :}$
$n_{Al_2(SO_4)_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,2=0,1mol$
$=>m_{Al_2(SO_4)_3}=0,1.342=34,2g$
$m_{dd\ spu}=5,4+784-0,3.2=788,8g$
$=>C\%_{Al_2(SO_4)_3}=\dfrac{34,2}{788,8}.100\%=4,34\%$
mddH2SO4=560.1,4=784g
=>mH2SO4=39,2g
=>nH2SO4=39,2\98=0,4 mol
=>nAl=5,4\27=0,2 mol
2Al+3H2SO4->Al2(SO4)3+3H2
ta có 0,2\2<0,4\3
=>nên H2SO4 dư
=>mH2SO4=0,2\98=19,6g
=>VH2=0,3.22,4=6,72l