\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(2Al+6HCl\rightarrow AlCl_3+3H_2\)
0,3--> 0,9------0,15--> 0,45
a)
\(m_{AlCl_3}=0,15.133,5=20,025\left(g\right)\)
b)
\(V_{H_2}=0,45.24,79=11,1555\left(l\right)\)
c) \(m_{HCl}=0,9.36,5=32,85\left(g\right)\)