$n_{Al}=\dfrac{5,4}{27}=0,2(mol)$
PTHH:
$2Al+6HCl\to 2AlCl_3+3H_2$
a,
Theo PTHH, $n_{HCl}=3n_{Al}=0,6(mol)$
$\to m_{HCl}=0,6.36,5=21,9g$
b,
Theo PTHH, $n_{H_2}=\dfrac{3}{2}n_{Al}=0,3(mol)$
$\to V_{H_2}=0,3.22,4=6,72l$
c,
+ Cách 1:
Theo PTHH, $n_{AlCl_3}=n_{Al}=0,2(mol)$
$\to m_{AlCl_3}=0,2.133,5=26,7g$
+ Cách 2:
$m_{H_2}=0,3.2=0,6g$
BTKL:
$m_{AlCl_3}=m_{Al}+m_{HCl}-m_{H_2}=5,4+21,9-0,,6=26,7g$