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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