A copper refining operation uses 34.4 mol of oxygen every second in the refining of copper (I) sulfide. Determine the number of moles of copper(I) oxide produced by this reaction every second.

Answer 1

Short answer: 22.9 mol of Cu2O per second.

The equation did not survive into your post, so start by writing it. Roasting copper(I) sulfide in oxygen to give copper(I) oxide is:

  2 Cu2S  +  3 O2  ->  2 Cu2O  +  2 SO2

Balance check, left against right: Cu 4 and 4, S 2 and 2, O 6 on the left against 2 from the two Cu2O plus 4 from the two SO2. Do not skip this step when you have supplied the equation yourself rather than being handed it.

The ratio that matters is Cu2O : O2 = 2 : 3.

  34.4 mol O2     2 mol Cu2O
  -----------  x  ----------  =  22.9 mol Cu2O per second
    1 second      3 mol O2

34.4 x 2/3 = 22.933..., so report 22.9 mol/s to three significant figures.

Why the rate wording is harmless. Everything in the problem is per second, so the time unit divides out and you are left doing ordinary stoichiometry. A steady-state process like this consumes and produces in fixed proportion every second, which is exactly what the coefficients describe.

The sulfur is where students lose points. It is tempting to think the oxygen all ends up in the copper(I) oxide, but two thirds of it leaves as SO2. That sulfur dioxide is not a footnote either: capturing it and converting it to sulfuric acid is a large part of what a real copper smelter does, and letting it escape is what caused the acid rain problems these plants were built to stop.

A quick reasonableness test. Cu2O carries a coefficient of 2 against O2 at 3, so the product count has to be smaller than 34.4. If you got 51.6 you inverted the ratio; if you got 68.8 you used the two Cu2O against a single O2.

Worth knowing where this sits in the process: the copper(I) oxide you just calculated is not the end point. It reacts with more of the sulfide in the next step, 2 Cu2O + Cu2S -> 6 Cu + SO2, and that is the step that finally frees the metal.