\(Fe2O3+3CO-->2Fe+3CO2\)
x--------3x(mol)--------------2x(mol)
\(CuO+CO-->Cu+H2O\)
y--------y-------------------y(mol)
\(n_{CO}=\frac{1,792}{22,4}=0,08\left(mol\right)\)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}160x+80y=4,8\\3x+y=0,08\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,02\end{matrix}\right.\)
\(m_{Fe}=0,02.2=56=2,24\left(g\right)\)
\(m_{Cu}=0,02.64=1,28\left(g\right)\)
\(m_{_{ }KL}=2,24+1,28=3,52\left(g\right)\)
Fe2O3+3CO-->2Fe+3CO2
x-------3x
CuO+CO-->Cu+CO2
y-------y
ta có
160x+80=4,8
3x+y=0,08
=>x=0,02 mol==>mFe2O3=0,02.160=3,2 g
y=0,02 mol
==>mCuo=4,8-3,2=1,6 g
