$m_{Al}=49,1\%.22=10,802g \\⇒m_{Fe}=22-10,802=11,198g \\n_{Al}=\dfrac{10,802}{27}≈0,4mol \\ n_{Fe}=\dfrac{11,198}{56}≈0,2mol \\PTHH : \\2Al+6HCl\to 2AlCl_3+3H_2↑(1) \\Fe+2HCl\to FeCl_2+H_2↑(2) \\a.Theo\ pt\ (1)\ và\ (2) : \\∑n_{HCl}=n_{HCl(1)}+n_{HCl(2)}=3.n_{Al}+2.n_{Fe}=3.0,4+2.0,2=1,6mol \\⇒m_{HCl}=1,6.36,5=58,4g \\b.Theo\ pt\ (1)\ và\ (2) : \\n_{H_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.1,6=0,8mol \\⇒V_{H_2}=0,8.22,4=17,92l \\c.n_{CuO}=\dfrac{72}{80}=0,9mol \\PTHH :$
$CuO + H_2\overset{t^o}\longrightarrow Cu+H_2O$
Theo pt : 1 mol 1 mol
Theo đbài : 0,9 mol 0,8 mol
Tỉ lệ : $\dfrac{0,9}{1}>\dfrac{0,8}{1}$
⇒Sau phản ứng CuO dư
$Theo\ pt : \\n_{CuO\ pư}=n_{H_2}=0,8 mol \\⇒n_{CuO\ dư}=0,9-0,8=0,1mol \\⇒m_{CuO\ dư}=0,1.80=8g \\n_{Cu}=n_{H_2}=0,8mol \\⇒m_{Cu}=0,8.64=51,2g \\⇒m_{rắn}=8+51,2=59,2g$
