What is the theoretical yield in grams of copper (ii) oxide when 47.77g of copper (ii) nitrate is reacted with 11.88g aluminum oxide?

The theoretical yield of copper (II) oxide is 20.20 grams

Explanation:

To calculate the number of moles, we use the equation:

     .....(1)

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:

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

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

the answer is c. nacl