$n_{Al}=\dfrac{5,4}{27}=0,2mol \\m_{HCl}=300.18,25\%=54,75g \\⇒n_{HCl}=\dfrac{54,75}{36,5}=1,5mol \\a.PTHH :$
$2Al + 6HCl\to 2AlCl_3+3H_2(1)$
Theo pt : 2 mol 6 mol
Theo đbài : 0,2 mol 1,5 mol
Tỉ lệ :$ \dfrac{0,2}{2}<\dfrac{1,5}{6}$
⇒Sau phản ứng HCl dư
$n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3mol \\⇒V_{H_2}=0,3.22,4=6,72l \\b.Theo\ pt\ (1): \\n_{HCl\ pư}=3.n_{Al}=3.0,2=0,6mol \\⇒n_{HCl\ dư}=1,5-0,6=0,9mol \\⇒m_{HCl\ dư}=0,9.36,5=32,85g \\n_{AlCl_3}=n_{Al}=0,2mol \\⇒m_{AlCl_3}=0,2.133,5=26,7g \\m_{H_2}=0,3.2=0,6g \\m_{dd\ sau\ phản\ ứng}=5,4+300-0,6=304,8g \\⇒C\%_{HCl\ dư}=\dfrac{32,85}{304,8}.100\%=10,77\% \\C\%_{AlCL_3}=\dfrac{26,7}{304,8}.100\%=8,76\% \\c.PTHH : \\Fe_2O_3+3H_2\overset{t^o}\to 2Fe+3H_2O(2) \\Theo\ pt\ (2) : \\n_{Fe}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,3=0,2mol \\⇒m_{Fe}=0,2.56=11,2g$
