2 NH3 + 3 CuO (s) --> N2 (g) + 3 Cu (s) + 3 H2O (g)
You know it is a limiting reagent (LR) problem when amounts for BOTH reactants are given.
b. Determine the LR first. I do these the long way.
mols NH3 = g/molar mass = 18.1/17 = apprx 1.1 but that's a close estimate. You need to redo ALL of the calculations because I've estimated all of them.
So how much N2 would this produce IF you had an excess of CuO. That will be 1.1 mols NH3 x (1 mol N2/2 mol NH3) = about 0.55
Now look at the CuO.
Mols CuO = 90.4/ about 80 = approx 1.2
How much N2 can be obtained from the CuO IF we had an excess of NH3. That's 1.2 mols CuO x (1 mol N2/3 mol CuO) = about 1.2/3 = about 0.4
In LR problem the smallest number is always the correct one so CuO is the LR.
How much N2 can be produced? That's 0.4 mols CuO x (1 mol N2/3 mols CuO) = about 0.4/3 = ? That is the theoretical yield of N2 in mols. Convert mols to grams for theoretical yield in grams.
Theoretical yield for Cu is done the same way.
Nitrogen gas can be prepared by passing ammonia over solid copper (II) oxide according to the equation
2 NH3 + 3 CuO (s) --> N2 (g) + 3 Cu (s) + 3 H2O (g)
Suppose 18.1 grams of ammonia is reacted with 90.4 grams CuO. a) What is the theoretical yield of nitrogen in grams?
g
b) What is the theoretical yield of copper in grams?
g
c) Which reactant is limiting?
2 answers
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