Đặt n_{Cu}=x(mol);n_{FeO}=y(mol);n_{Fe_2O_3}=z(mol)
->64x+72y+160z=46(1)
3Cu+8HNO_3->3Cu(NO_3)_2+2NO+4H_2O
3FeO+10HNO_3->3Fe(NO_3)_3+NO+5H_2O
Fe_2O_3+6HNO_3->2Fe(NO_3)_3+3H_2O
Theo PT:
n_{NO}=2/{3}x+1/{3}y=0,25(2)
n_{HNO_3}=8/{3}x+{10}/{3}y+6z=1,9(3)
(1)(2)(3)->x=0,3;y=0,15;z=0,1
Theo PT: n_{Cu(NO_3)_2}=x=0,3(mol);n_{Fe(NO_3)_3}=y+2z=0,35(mol)
->m=0,3.188+0,35.242=141,1(g)
\%m_{Cu}={0,3.64}/{46}.100\%\approx 41,74\%
\%m_{FeO}={0,15.72}/{46}.100\%\approx 23,48\%
\%m_{Fe_2O_3}=100-23,48-41,74=34,78\%
