The theoretical yield of copper (II) oxide is 20.20 grams
Explanation:
To calculate the number of moles, we use the equation:
.....(1)
For Copper (II) nitrate:
Given mass of copper (II) nitrate = 47.77 g
Molar mass of copper (II) nitrate = 187.56 g/mol
Putting values in equation 1, we get:
For Aluminum oxide:
Given mass of aluminum oxide = 11.88 g
Molar mass of aluminum oxide = 102 g/mol
Putting values in equation 1, we get:
The chemical equation for the formation of ammonia follows:
By Stoichiometry of the reaction:
3 moles of copper (II) nitrate reacts with 1 mole of aluminum oxide
So, 0.254 moles of copper (II) nitrate will react with = 
of aluminum oxide
As, given amount of aluminum oxide is more than the required amount. So, it is considered as an excess reagent.
Thus, copper (II) nitrate is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
3 moles of copper (II) nitrate produces 3 moles of copper (II) oxide
So, 0.254 moles of copper (II) nitrate will produce = 
of copper (II) oxide
Now, calculating the theoretical yield of copper (II) oxide from equation 1, we get:
Molar mass of copper (II) oxide = 79.545 g/mol
Moles of copper (II) oxide = 0.254 moles
Putting values in equation 1, we get:
Hence, the theoretical yield of copper (II) oxide is 20.20 grams
