Chemistry Homework Help Please!

Question

Reacting mixture of 1.000 kg of each of the reagents are used.

aluminum copper(ll) oxide equation:

2Al + CuO ---> Al₂O + Cu

1. what is the limiting reaction and how much remains theoretically unused?
2. what is the excess reaction and how much remains theoretically unused?
3. How many kilograms of metal are produced?
4. what is the oxidizing and reducing reaction? why?

Answer 1

Well, unless you were given that equation, it is incorrect. The formula for aluminum oxide is Al₂O₃, not Al₂O. And then the balanced equation would also be incorrect & should be 2 Al + 3 CuO → Al₂O₃ + 3 Cu

Assuming this IS the correct equation, we can now solve the problem correctly.

I would solve #2 first, which will then answer #1 as well.

To calculate the theoretical yield of Cu (the metal) we will calculate the theoretical yield twice: once as if the Al is the limiting reactant & then as if the CuO is the LR. The smaller result IS the theoretical yield & whatever produced that is the LR.

Using the Al: (63.54 g Cu/1 mol Cu) (3 mol Cu/2 mol Al) (1 mol Al/27.0 g Al) 1000 g Al = ~ 3530 g = 3.50 kg Cu

Using the CuO: (63.54 g Cu/1 mol Cu) (3 mol Cu/3 mol CuO) (1 mol CuO/79.54 g CuO) 1000 g Cu) = ~ 799 g = 0.799 kg Cu.

Therefore, the TY of Cu is 0.759 kg and the CuO is the LR.

Now, to calculate the amount of the XS reactant that is left over, we can use the amount of CuO that reacted & see how much of the Al actually reacted. Then we take the difference between what reacted & that available to get the XS.

(27.0 g Al/1 mol Al) (2 mol Al/3 mol CuO) (1 mol CuO/79.54 g CuO) 1000 g CuO = ~ 226 g = 0.226 kg Al reacted.

1.000 kg - 0.226 kg = 0.774 kg Al in XS.

The Al went from a charge of 0 to 3+, so it lost 3 electrons. Thus, oxidation. The Cu went from 2+ to 0, gaining 2 electrons: reduction.