Bài 5:

Đặt $n_{FeO}=x(mol);n_{Fe_2O_3}=y(mol)$

$\Rightarrow 72x+160y=24,1(1)$

$n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8(mol)$

$a,PTHH:FeO+2HCl\to FeCl_2+H_2O$

$Fe_2O_3+6HCl\to 2FeCl_3+3H_2O$

$\Rightarrow 2x+6y=0,8(2)$

$(1)(2)\Rightarrow x=0,15(mol);y=0,08(mol)$

$b,\%m_{FeO}=\dfrac{0,15.72}{24,1}.100\%=44,81\%$

$\%m_{Fe_2O_3}=100\%-44,81\%=55,19\%$

$c,n_{FeCl_2}=x=0,15(mol);n_{FeCl_3}=2y=0,16(mol)$

$\Rightarrow m_{FeCl_2}=0,15.127=19,05(g)$

$m_{FeCl_3}=0,16.162,5=26(g)$

$\Rightarrow C\%_{FeCl_2}=\dfrac{19,05}{0,15.72+200}.100\%=9,04\%$

$C\%_{FeCl_3}=\dfrac{26}{0,08.160+200}.100\%=12,22\%$

Bài 6:

Đặt $n_{Al}=x(mol);n_{Mg}=y(mol)$

$\Rightarrow 27x+24y=7,8(1)$

$a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2$

$Mg+2HCl\to MgCl_2+H_2$

$\Rightarrow n_{H_2}=1,5x+y$

$\Rightarrow 7,8+500-2(1,5x+y)=507$

$\Rightarrow 1,5x+y=0,4(2)$

$(1)(2)\Rightarrow x=0,2(mol);y=0,1(mol)$

$\Rightarrow m_{Al}=0,2.27=5,4(g);m_{Mg}=0,1.24=2,4(g)$

$b,\Sigma n_{HCl}=3x+2y=0,6+0,2=0,8(mol)$

$\Rightarrow C\%_{HCl}=\dfrac{0,8.36,5}{500}.100\%=5,84\%$