Al + 2HCl → AlCl\(_2\) +H\(_2\)↑
+n\(_{Al}=\frac{5,4}{27}=0,2\)(mol)
+\(n_{H2}=n_{Al}\)=0,2(mol)
+\(V_{H2}\)=n.22,4=0,2.22,4=4,48(lit)
+n\(AlCl_2\)=n\(_{Al}\)=0,2(mol)
+\(m_{AlCl2}\)=n.M=0,2.(27.35,5.2)=19,6(gam)
Theo DLBTKL ta có:
+m\(_{dungdich}\)= m\(_{HCl}-m_{Al}+m_{H2}\)=75-5,4+(0,2.2)=70(g)
C%=\(\frac{m_{châttan}}{m_{dugndich}}\).100%=\(\frac{19,6}{70}.100\%\)=28%