ANSWERS: 1
  • Dude did you try solving it urself? Mr (NH3) = 14 + 1 + 1 + 1 = 17 Mr (CuO) = 63.5 + 16 = 79.5 No. of mol of NH3 used = 18.1/17 = 1.06471 (approx) No. of mol of CuO used = 90.4/79.5 = 1.1371 (approx) Since the mol ratio of NH3 : CuO = 2:3, More CuO was used in the reaction, therefore by comparison, CuO was the limiting agent in this equation; NH3 was in excess. --- This step is not needed in your answer This basically means that all the CuO reacted, and that some NH3 remains. No. of mol of NH3 reacted = 1.1371 x 2 / 3 = 0.7581 (approx) If ALL the NH3 had reacted, you would have needed 1.59706 mol of CuO which is more than the max amount of CuO which you have. ---- Since the mol ratio of CuO : N2 = 3:1, No. of mol of N2 produced = 1.1371 / 3 = 0.3790 (approx) Mr (N2) = 14 + 14 = 28 Therefore mass of N2 produced = 28 x 0.3790 = 10.612 g = 10.6 g (3 sf)

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