\(n_{Al}=\dfrac{1.35}{27}=0,05\left(mol\right)\); \(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
Xét tỉ lệ: \(\dfrac{0,05}{2}< \dfrac{0,2}{6}\) => Al hết, HCl dư
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
_____0,05------------------------>0,075
=> V = 0,075.22,4 = 1,68(l)
=> B