Question

Answer 1

\(n_{Al}=\dfrac{8,1}{27}=0,3mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Theo pt: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45mol\Rightarrow V_{H_2}=10,08l\)

\(n_{HCl}=3n_{Al}=0,9mol\Rightarrow m_{HCl}=32,85g\)

\(n_{AlCl_3}=n_{Al}=0,3mol\Rightarrow m_{AlCl_3}=40,05g\)