2Al + 6HCl --> 2AlCl3 + 3H2
0,15 0,45 0,15 0,225 (mol)
\(n_{H_2}=\dfrac{5,04}{22,4}=0,225mol\)
a. \(m_{HCl}=0,45.36,5=16,425g\)
\(\%HCl=\dfrac{16,425.100}{296.4}=5,54\%\)
b. \(m_{AlCl_3}=0,15.133,5=20,025g\)
\(m_{dd_{AlCl_3}}=m_{Al}+m_{dd_{HCl}}-m_{H_2}\)
\(=4,05+296,4-0,45=300g\)
\(\%AlCl_3=\dfrac{20,025.100}{300}=6,675\%\)