$n_{H_2SO_4}=1,35.0,4=0,54mol$
$n_{Al}=10/27≈0,37mol$
$a.$ $2Al + 3H_2SO_4\to Al_2(SO_4)_3+3H_2↑$
b.Theo pt : 2 mol 3 mol
Theo đbài : 0,37 mol 0,54 mol
Tỷ lệ : $\dfrac{0,37}{2}>\dfrac{0,54}{3}$
⇒Sau pư Al dư
Theo pt :
$n_{Al\ pư}=2/3.n_{H_2SO_4}=2/3.0,54=0,36mol$
$⇒n_{Al\ dư}=0,37-0,36=0,01mol$
$⇒m_{Al\ dư}=0,01.27=0,27g$
c.Theo pt :
$n_{H_2}=n_{H_2SO_4}=0,54mol$
$⇒V_{H_2}=0,54.22,4=12,096l$
d.Theo pt :
$n_{Al_2(SO_4)_3}=1/3.n_{H_2SO_4}=1/3.0,54=0,18mol$
$⇒C_{M_{Al_2(SO_4)_3}}=\dfrac{0,18}{0,4}=0,45M$