a. 2Al + 6HCl -> 2AlCl3 + 3H2
b. nAl = \(\dfrac{8.1}{27}=0,3\left(mol\right)\)=> \(n_{H_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\)
\(V_{H_2}=0,45.22,4=10,08\left(mol\right)\)
c. \(n_{HCl}=3n_{Al}=3.0,3=0,9\left(mol\right)=>m_{HCl}=0,9.36,5=32,85\left(g\right)\)
Vậy m = 32,85
d. C1: Dùng ĐLBTKL: \(m_{AlCl_3}=8,1+32,85-0,45.2=40,05\left(g\right)\)
C2: \(n_{AlCl_3}=n_{Al}=0,3mol=>m_{AlCl_3}=0,3.98=40,05\left(g\right)\)