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AS Chemistry query

Here's the question: A passenger jet contains 4050 kg of copper wiring.
A rock sample contains 1.25% CuFeS2 by mass.
Calculate the mass, in tonnes, of rock needed to produce enough copper wire for
a passenger jet. (1 tonne = 1000 kg)

The previous part of the question (6.1) asked to balance the equations:
4CuFeS2 + 9/2 O2 + 4SiO2 Cu2S + Cu2O + 7SO2 + 4FeSiO3
Cu2S + 2Cu2O 6Cu + SO2

My thought process was, calc moles of Cu in 4050kg, which is 4,050,000/63.5= 63,779....

Then in the balanced equations 6 moles of Cu originally comes from 4 moles CuFeS2, so multiply these moles by 4/6, to get 42,519.... moles CuFeS2. Calculate the mass, this would be molesxMr=42,519...*183.5
=7,802,362.205g =7802.362205kg.
In the rock, 1.25% = CuFeS2, so multiply by 100/1.25, then divide by 100 to get answer in tonnes.

I got: 624.1889764 tonnes. The correct answer was 936 tonnes, and they used the original 63,779... moles to calculate the mass of CuFeS2, I don't understand why and if anyone could help out that would be greatly appreciated.
Original post by harryxdx2
Here's the question: A passenger jet contains 4050 kg of copper wiring.
A rock sample contains 1.25% CuFeS2 by mass.
Calculate the mass, in tonnes, of rock needed to produce enough copper wire for
a passenger jet. (1 tonne = 1000 kg)

The previous part of the question (6.1) asked to balance the equations:
4CuFeS2 + 9/2 O2 + 4SiO2 Cu2S + Cu2O + 7SO2 + 4FeSiO3
Cu2S + 2Cu2O 6Cu + SO2

My thought process was, calc moles of Cu in 4050kg, which is 4,050,000/63.5= 63,779....

Then in the balanced equations 6 moles of Cu originally comes from 4 moles CuFeS2, so multiply these moles by 4/6, to get 42,519.... moles CuFeS2. Calculate the mass, this would be molesxMr=42,519...*183.5
=7,802,362.205g =7802.362205kg.
In the rock, 1.25% = CuFeS2, so multiply by 100/1.25, then divide by 100 to get answer in tonnes.

I got: 624.1889764 tonnes. The correct answer was 936 tonnes, and they used the original 63,779... moles to calculate the mass of CuFeS2, I don't understand why and if anyone could help out that would be greatly appreciated.


“Then in the balanced equations 6 moles of Cu originally comes from 4 moles CuFeS2” - this does not obey the law of conservation of mass, as it implies that 2 moles of copper atoms appeared from nowhere. This is a big hint that you’ve used an incorrect ratio.

I don’t think you’ve balanced the first equation correctly - firstly, look at the number of oxygen atoms on either side and secondly, try a different initial number of moles of CuFeS2.

I balanced the first equation differently and so, I got an equation wherein the right-hand side had 1 mole of Cu2S and 2 moles of Cu2O on it, among the other products.

With the first equation as I balanced it, I followed essentially the same method as you outlined and reached the same answer as the MS.
(edited 11 months ago)
Reply 2
Original post by TypicalNerd
“Then in the balanced equations 6 moles of Cu originally comes from 4 moles CuFeS2” - this does not obey the law of conservation of mass, as it implies that 2 moles of copper atoms appeared from nowhere. This is a big hint that you’ve used an incorrect ratio.

I don’t think you’ve balanced the first equation correctly - firstly, look at the number of oxygen atoms on either side and secondly, try a different initial number of moles of CuFeS2.

I balanced the first equation differently and so, I got an equation wherein the right-hand side had 1 mole of Cu2S and 2 moles of Cu2O on it, among the other products.

With the first equation as I balanced it, I followed essentially the same method as you outlined and reached the same answer as the MS.

OHHHHHHHH thank you very much, no clue how I didn't spot this one to be honest.

Appreciate it!
Original post by harryxdx2
OHHHHHHHH thank you very much, no clue how I didn't spot this one to be honest.

Appreciate it!


No problem. In fairness, they did give you an equation that can be balanced in an infinite number of ways, given it is a combination of multiple reactions

The trick was to balance the second equation first, as it is simpler and then make sure that what appears on the left hand side of the second equation appears, in the exact same quantities, on the right hand side of the first.
(edited 11 months ago)

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