Question

What is the limiting reactant when 19.9g of CuO are exposed to 2.02g of H2 according to the following equation?

CuO + H2 ---> Cu + H20



My attempt at this:

19.9gCuO * 1moleCuO/79.539gCuO * 1moleCu/1moleCuO = .250moleCu

2.02gH2 * 1moleH2/2.016gH2 * 1moleCu/1moleH2 = 1.00moleCu

The first one is limiting reactant?

B. How many grams of Cu are produced?

.250moleCu * 63.54gCu/1moleCu = 15.9gCu


Is that correct?

Answer 1

Both A and B look ok to me.

Answer 2

Thanks!