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) sulfide required for this reaction every second.

Answer 1

Short answer: 22.9 mol of Cu2S per second.

Your post does not print the equation, so the first job is to pin it down. Roasting copper(I) sulfide with copper(I) oxide as the product balances like this:

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

  Cu:   4              4
  S:    2              2
  O:         6         2 + 4 = 6

I can be sure it is that reaction and not the smelting one (Cu2S + O2 -> 2 Cu + SO2) because the companion part of this same problem asks for moles of copper(I) oxide produced. When a reaction is missing from the prompt, let the named products pin it down for you.

Now the arithmetic. The only ratio you need is Cu2S : O2 = 2 : 3.

  34.4 mol O2     2 mol Cu2S
  -----------  x  ----------  =  22.9 mol Cu2S per second
    1 second      3 mol O2

34.4 x 2/3 = 22.933..., which rounds to 22.9 mol/s, three significant figures to match the 34.4 you were given.

Trap 1: "per second" is not a complication. A rate ratio behaves exactly like a mole ratio, because the seconds sit on both sides and cancel. If the wording bothers you, solve the problem for a plain 34.4 mol of oxygen and staple "per second" back on at the end. Nothing about the stoichiometry changes.

Trap 2: check the direction before you trust the number. The coefficient on Cu2S (2) is smaller than the one on O2 (3), so your answer must land below 34.4. It does. That one glance catches the single most common error here, which is multiplying by 3/2 instead of 2/3 and reporting 51.6.

One bonus, since it is the other half of your problem set: the copper(I) oxide comes out at 22.9 mol/s as well. Cu2S and Cu2O both carry a coefficient of 2, so one conversion answers both questions.