FeO+CO  $$\xrightarrow{t^o}$$  Fe+CO_2

Fe_2O_3+3CO  $$\xrightarrow{t^o}$$  2Fe+3CO_2

Fe_3O_4+4CO  $$\xrightarrow{t^o}$$  3Fe+4CO_2

CuO+CO  $$\xrightarrow{t^o}$$  Cu+CO_2

Theo PT: n_{CO}=n_{CO_2}=n_{O(o x i t)}

BTKL: m_{O(o x i t)}=m_X-m_Y=a+1,92-a=1,92(g)

->n_O={1,92}/{16}=0,12(mol)

FeO+2HCl->FeCl_2+H_2O

Fe_2O_3+6HCl->2FeCl_3+3H_2O

Fe_3O_4+8HCl->2FeCl_3+FeCl_2+4H_2O

CuO+2HCl->CuCl_2+H_2O

CuCl_2+2AgNO_3->Cu(NO_3)_2+2AgCl

FeCl_2+3AgNO_3->Fe(NO_3)_3+2AgCl+Ag

FeCl_3+3AgNO_3->Fe(NO_3)_3+3AgCl

Theo PT: n_{HCl}=2n_O=0,24(mol);n_{H_2O}=n_O=0,12(mol)

BTKL: a+1,92+0,24.36,5=2a+2,6+0,12.18

->a=5,92(g)

Bảo toàn Cl: n_{AgCl}=n_{HCl}=0,24(mol)

->m_{Ag}=6.5,92+1,08-0,24.143,5=2,16(g)

->n_{Ag}={2,16}/{108}=0,02(mol)

Theo PT: n_{FeCl_2}=n_{Ag}=0,02(mol)

Đặt n_{FeCl_3}=x(mol);n_{CuCl_2}=y(mol)

Theo PT: 3x+2y+2n_{FeCl_2}=n_{AgCl}

->3x+2y=0,2(1)

Lại có 162,5x+135y+0,02.127=2.5,92+2,6(2)

(1)(2)->x=0,04(mol);y=0,04(mol)

Bảo toàn Fe: n_{Fe(Y)}=x+n_{FeCl_2}=0,06(mol)

Bảo toàn Cu: n_{Cu(Y)}=y=0,04(mol)

->\%m_{Fe}={0,06.56}/{0,06.56+0,04.64}.100\%\approx 56,76\%

->\%m_{Cu}=100-56,76=43,24\%