a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, Ta có: \(n_{H_2}=0,12\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,24\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,24}{0,5}=0,48\left(M\right)\)
c, Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,08\left(mol\right)\)
\(\Rightarrow m_{Al}=0,08.27=2,16\left(g\right)\)
d, Theo PT: \(n_{AlCl_3}=\dfrac{2}{3}n_{H_2}=0,08\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,08.133,5=10,68\left(g\right)\)
\(C_{M_{AlCl_3}}=\dfrac{0,08}{0,5}=0,16\left(M\right)\)
Bạn tham khảo nhé!