Copper Stoichometry Ratio Problem I was given two equations as part of a problem into Copper Wiring production for planes:

$$\ce{4CuFeS2 + 9 1/2 O2 + 4SiO2 \rightarrow Cu2S + Cu2O + 2SO2 + 4FeSiO3}$$
$$\ce{Cu2S + 2Cu2O \rightarrow 6Cu + SO2}$$

This question later became the source of a Chemistry exam question for the English IB equivalent. 

I was given the following information. 


  The previous equations are part of the manufacture of copper for wiring in planes. 
  A passenger jet contains $4050\ \pu{ kg}$ of copper wiring.
  A rock sample contains $1.25$% $\ce{CuFeS2}$ by mass.
  Calculate the mass, in tonnes, of rock needed to produce enough copper wire for a passenger jet. ($\pu{1~tonne} = \pu{1000~kg}$)


Now, I tried to solve it as follows:

Adding both equations gave:

$$\ce{4CuFeS2 + 9 1/2 O2 + 4SiO2 + Cu2O \rightarrow 2SO2 + 4FeSiO3 + 6Cu + SO2}$$

Now, I can see that $4$ moles of $\ce{CuFeS2}$ from the ore would give us the required $\ce{6Cu}$ for the production.

However, the solution states that Cu and $\ce{CuFeS2}$ are in a $1:1$ ratio, which I couldn't explain why. 



Using the equation I formed above, this idea didn't make sense. 
Why is the ratio of Cu to $\ce{4CuFeS2}$ $1:1$?
No serious mining operation would just dump a third of their desired product. 
 The question does say that the reactions are "part of the manufacture of copper." This does not preclude other reactions. With the given reactions alone we are not able to handle the surplus of $\ce{Cu2S}$.
$$\ce{\color{\red}{4CuFeS2} + 9 1/2 O2 + 4 SiO2 \rightarrow \color{\red}{Cu2S + Cu2O} + 2SO2 + 4FeSiO3}  \tag{1}$$
$$\ce{\color{\red}{2} Cu2S + 2 Cu2O \rightarrow 6Cu + SO2 +\color{\red}{Cu2S}}\tag{2}$$
However we can infer from the first reaction that the sulfide can react with oxygen to form the needed oxide to balance reduction and oxidation.
$$\ce{ Cu2S + 1 1/2O2 -&gt;  Cu2O + SO2}\tag{3}$$
The second equation can now become:
$$\ce{\color{\red}{2} Cu2S + 2 Cu2O +\color{\red}{1 1/2 O2} \rightarrow 8 Cu + 2 SO2 \tag{4}}$$
Thus in the first step $\ce{Cu2O \text{and} Cu2S}$ are produced according to quation 1.  In the second step they are reduced with aditional oxygen according to equation 4. To pove they balance first double equation 1.
$$\ce{8CuFeS2 + 19 O2 + 8 SiO2 \rightarrow \color{\red}{2Cu2S + 2Cu2O} + 4 SO2 + 8 FeSiO3}  \tag{5.1}$$
Add $\ce{1 1/2 O2}$ to each side
$$\ce{8CuFeS2 + \color{\red}{20 1/2 O2} + 8 SiO2 \rightarrow \color{\red}{2Cu2S + 2Cu2O + 1 1/2 O2 }+ 4 SO2  + 8 FeSiO3}  \tag{5.2}$$
Substitute the red products with equation 4 reactants.
$$\ce{8CuFeS2 + \color{\red}{20 1/2 O2} + 8 SiO2 \rightarrow \color{\red}{8 Cu + 2 SO2}+ 4 SO2  + 8 FeSiO3}  \tag{5.3}$$
This Gives a final balanced equation of 
$$\ce{8CuFeS2 + 20 1/2 O2 + 8 SiO2 \rightarrow 8 Cu + 6 SO2  + 8 FeSiO3}  \tag{5.4}$$
As you can see on a molar eatio $\ce{CuFeS2\text{:}Cu}$ is 1:1 after accounting for the additional oxygen needed.