Asked by Chopsticks
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?
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?
Answers
Answered by
DrBob222
Both A and B look ok to me.
Answered by
Chopsticks
Thanks!
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