$n_{Al}=10,8/27=0,4mol$
$PTHH :$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑$
$\text{a.Theo pt :}$
$n_{Al_2(SO_4)_3}=1/2.n_{Al}=1/2.0,4=0,2mol$
$⇒m_{Al_2(SO_4)_3}=0,2.342=68,4g$
$\text{b.Theo pt :}$
$n_{H_2}=3/2.n_{Al}=3/2.0,4=0,6mol$
$⇒V_{H_2}=0,6.22,4=13,44l$
$\text{c.Nếu H2 hao hụt 10%}$
$⇒V_{H_2\ tt}=90\%.13,44=12,096l$
$\text{d.Giả sử khí H2 bị hao hụt 25%}$
$⇒V_{H_2\ tt}=125\%.11,2=14l$
$⇒n_{H_2}=14/22,4=0,625mol$
$\text{Theo pt :}$
$n_{Al}=2/3.n_{H_2}=2/3.0,625=\dfrac{5}{12}mol$
$⇒m_{Al}=\dfrac{5}{12}.27=11,25g$