Bài 1 :
$PTHH :$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑$
$n_{Al_2(SO_4)_3}=34,2/342=0,1mol$
$Theo\ pt : $
$⇒n_{Al}=2n_{Al_2(SO_4)_3}=2.0,1=0,2mol$
$⇒m_{Al}=0,2.27=5,4gam$
$n_{H_2}=3/2.n_{Al}=3/2.0,2=0,3mol$
$⇒V_{H_2}=0,2.22,4=6,72l$
Bài 2 :
$n_{Al}=5,4/27=0,2mol$
$2Al+6HCl\to 2AlCl_3+3H_2↑$
$Theo\ pt : $
$n_{H_2}=3/2.n_{Al}=3/2.0,2=0,3mol$
$⇒V_{H_2}=0,3.22,4=6,72l$
$n_{AlCl_3}=n_{Al}=0,2mol$
$⇒m_{AlCl_3}=0,2.133,5=26,7g$