\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Theo PTHH : \(n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2}=n.M=0,3.2=0,6\left(mol\right)\)
\(m_{ddAlCl_3}=m_{Al}+m_{ddHCl}-m_{H_2}\)
\(=5,4+120-0,6=124,8\left(g\right)\)
Theo PTHH: \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=n.M=0,2.133,5=26,7\left(g\right)\)
\(C_{\%AlCl_3}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{26,7}{124,8}.100\%=21,4\%\)
\(V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
2Al+6HCl\(\rightarrow2AlCl_3+3H_2\)
\(n_{HCl}=3n_{Al}=0,6mol\rightarrow m_{HCl}=0,6x36,5=21,9g\)
\(C\%_{HCl}=\dfrac{21,9}{120}.100=18,25\%\)
\(n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,2=0,3mol\rightarrow V_{H_2}=0,3x22,4=6,72l\)
\(n_{AlCl_3}=n_{Al}=0,2mol\rightarrow m_{AlCl_3}=0,2x133,5=26,7g\)
mdd=mAl+mdd HCl-mH2=5,4+120-0,3x2=124,8g
\(C\%_{AlCl_3}=\dfrac{26,7}{124,8}.100\approx21,4\%\)
