\(n_{H_2}=\dfrac{20,16}{22,4}=0,9\left(mol\right)\)
PTHH :
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,6 1,8 0,6 0,9
\(a,m_{HCl}=1,8.36,5=65,7\left(g\right)\)
\(C\%_{HCl}=\dfrac{65,7}{400}.100\%=16,425\%\)
\(b,m_{AlCl_3}=0,6.133,5=80,1\left(g\right)\)
\(m_{ddAlCl_3}=\left(0,6.27+400\right)-0,9.2=414,4\left(g\right)\)
\(C\%_{AlCl_3}=\dfrac{80,1}{414,4}.100\%\approx19,33\%\)