nSO2 = 0,25 mol
Đặt nCu = x ; nFe = y ( x, y > 0 )
Cu + 2H2SO4 (đặc) \(\underrightarrow{t^o}\) CuSO4 + SO2 + 2H2O
x.....................................................x
2Fe + 6H2SO4(đặc) \(\underrightarrow{t^o}\) Fe2(SO4)3 + 3SO2 + 6H2O
y.........................................................1,5y
Ta có hệ
\(\left\{{}\begin{matrix}64x+56y=12\\x+1,5y=0,25\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\Rightarrow\) %Cu = \(\dfrac{0,1.64.100\%}{12}\)\(\approx\) 53,3%
\(\Rightarrow\) %Fe = \(\dfrac{0,1.56.100\%}{12}\) \(\approx\) 46,7%
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