a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
Ta có: \(m_{HCl}=200.21,9\%=43,8\left(g\right)\Rightarrow n_{HCl}=\dfrac{43,8}{36,5}=1,2\left(mol\right)\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)
\(n_{HCl}=3n_{Al}+2n_{MgO}=1,2\Rightarrow n_{MgO}=0,45\left(mol\right)\)
\(\Rightarrow x=m_{Al}+m_{MgO}=0,1.27+0,45.40=20,7\left(g\right)\)
b, Ta có: m dd sau pư = 20,7 + 200 - 0,15.2 = 220,4 (g)
Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\\n_{MgCl_2}=n_{MgO}=0,45\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{AlCl_3}=\dfrac{0,1.133,5}{220,4}.100\%\approx6,06\%\\C\%_{MgCl_2}=\dfrac{0,45.95}{220,4}.100\%\approx19,4\%\end{matrix}\right.\)
