2Al + 6HCl → 2AlCl3 + 3H2
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}\times0,3=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}=0,2\times27=5,4\left(g\right)\)
Theo PT: \(n_{HCl}=2n_{H_2}=2\times0,3=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,6\times36,5=21,9\left(g\right)\)
Theo PT: \(n_{AlCl_3}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}\times0,3=0,2\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,2\times133,5=26,7\left(g\right)\)
\(2Al+6HCl-->2AlCl_3+3H_2\)
0,2____0,6_______0,2_______0,3
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
=>\(m_{Al}=0,2.27=5,4\left(g\right)\)
=>\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)
\(m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\)
