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.

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Using the equation I formed above, this idea didn't make sense. Why is the ratio of Cu to $\ce{4CuFeS2}$ $1:1$?


3 Answers 3


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. $$\ce{ Cu2S + 1 1/2O2 -> Cu2O + SO2}\tag{3}$$ 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.


Suppose both reactions have gone to completion. Then, according to the first equation, $1$ moles of $\ce{CuFeS2}$ will produce $\frac{1}{4}$ moles of $\ce{Cu2S}$ and $\frac{1}{4}$ moles of $\ce{Cu2O}$. According to the second equation, $2$ moles of $\ce{Cu2O}$ and $1$ moles of $\ce{Cu2S}$ react together to produce $6$ moles of $\ce{Cu}$. Thus, $\frac{1}{4}$ moles of $\ce{Cu2O}$ and $\frac{1}{8}$ moles of $\ce{Cu2S}$ react together to produce $(6 \times \frac{1}{8}) =\frac{3}{4}$ moles of $\ce{Cu}$. Therefore, you need $\frac{4}{3}$ moles of $\ce{CuFeS2}$ to produce $1$ moles of $\ce{Cu}$. Thus, you need $\frac{4}{3} \times 63780$ moles of $\ce{CuFeS2}$ to produce $63780$ moles of $\ce{Cu}$ for the plane. Thus, if the rock consists solely of $\ce{Cu2S}$, mass of rock needed to produce this much $\ce{Cu}$ is:

$$\frac{4}{3} \times \pu{63780mol} \times \pu{183.5 \frac{g}{mol}} \times \frac{100}{1.25} \times \frac{\pu{1kg}}{\pu{1000 g}} \times \frac{\pu{1 tonne}}{\pu{1000 kg}} = \pu{1248 tonnes}$$


The problem I see here is the question has assumed that all of the $\ce{CuFeS2}$ will be converted into elemental copper at the end of the reaction. However whether or not this is possible also depends on the amount of $\ce{Cu2O}$ present.

If the rock consists solely of $\ce{CuFeS2}$, then the $x$ moles of $\ce{CuFeS2}$ will react to produce $x/4$ moles of $\ce{Cu2S}$ and $\ce{Cu2O}$. And then, $x/4$ moles of $\ce{Cu2O}$ will react with $x/8$ moles of $\ce{Cu2S}$ to produce $x/8$ moles of copper. Some of the copper compound $\ce{Cu2S}$ will be left unreacted because there is not enough $\ce{Cu2O}$ to react with it. As we can see, not all the copper compound present initially was converted to elemental copper. The answer has not considered the possibility that conversion of the copper compound may be incomplete.


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