\(m_{HCl}=\dfrac{C\%.m_{dd}}{100\%}=\dfrac{20\%.200}{100\%}=40\left(g\right)\)
=> \(n_{HCl}=\dfrac{m}{M}=\dfrac{40}{36,5}=1,1\left(mol\right)\)
\(n_{Al}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH : 2Al + 6HCl --> 2AlCl3 + 3H2
2 : 6 : 2 : 3
0.1 0.3 : 0.1 0,15
mol mol mol mol
Dễ thấy : \(\dfrac{n_{Al}}{2}< \dfrac{n_{HCl}}{6}\)
=> HCl dư 1,1 - 0,3 = 0,8 (mol)
\(m_{AlCl_3}=n.M=0,1.162,5=16,25\left(g\right)\)
\(m_{HCl\text{dư }}=n.M=0,8.98=78.4\left(g\right)\)
\(m_{H_2}=n.M=0,15.2=0,3\left(g\right)\)
mddmới = 5,6 + 200 - 0,3 = 205,3 (g)
=> \(C\%_{HCl}=\dfrac{m_{HCl}}{m_{dd\text{mới }}}.100\%=\dfrac{78.4}{205.3}.100\%=38,2\%\)
\(C\%_{AlCl_3}=\dfrac{m_{AlCl_3}}{m_{dd\text{mới }}}.100\%=\dfrac{16.25}{205.3}.100\%=8\%\)
