Question

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?

Answer 1

20.20 grams

Step-by-step explanation:

Answer 2

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

Step-by-step explanation:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(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:

`\text{Moles of copper (II) nitrate}=(47.77g)/(187.56g/mol)=0.254mol`

• 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:

`\text{Moles of aluminum oxide}=(11.88g)/(102g/mol)=0.12mol`

The chemical equation for the formation of ammonia follows:

`3Cu(NO_3)_2+Al_2O_3\rightarrow 2Al(NO_3)_3+3CuO`

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 =
`(1)/(3)* 0.254=0.084mol` 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 =
`(3)/(3)* 0.254=0.254moles` 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:

`0.254mol=\frac{\text{Mass of copper (II) oxide}}{79.545g/mol}\\\\\text{Mass of copper (II) oxide}=20.20g`

Hence, the theoretical yield of copper (II) oxide is 20.20 grams