\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
a. Theo PT ta có: \(n_{Al}=\dfrac{0,3.2}{3}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)
b. Theo PT ta có: \(n_{HCl}=\dfrac{0,3.6}{3}=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)
\(\Rightarrow C\%HCl=\dfrac{21,9}{200}.100\%=10,95\%\)