a. \(n_{Al}=\dfrac{5.4}{27}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{50.21,9\%}{36,5}=0,3\left(mol\right)\)
PTHH : 2Al + 6HCl -> 2AlCl3 + 3H2
0,1 0,3 0,1 0,15
Ta thấy : \(\dfrac{0.2}{2}>\dfrac{0.3}{6}\) => Al dư , HCl đủ
\(m_{dd}=5,4+50-\left(0,15.2\right)=55,1\left(g\right)\)
\(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
\(C\%_{AlCl_3}=\dfrac{13,35}{55,1}.100\%=24,22\%\)