Respuesta :
Answer: The theoretical yield of copper (II) oxide is 20.20 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(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:
[tex]\text{Moles of copper (II) nitrate}=\frac{47.77g}{187.56g/mol}=0.254mol[/tex]
- 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:
[tex]\text{Moles of aluminum oxide}=\frac{11.88g}{102g/mol}=0.12mol[/tex]
The chemical equation for the formation of ammonia follows:
[tex]3Cu(NO_3)_2+Al_2O_3\rightarrow 2Al(NO_3)_3+3CuO[/tex]
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 = [tex]\frac{1}{3}\times 0.254=0.084mol[/tex] 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 = [tex]\frac{3}{3}\times 0.254=0.254moles[/tex] 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:
[tex]0.254mol=\frac{\text{Mass of copper (II) oxide}}{79.545g/mol}\\\\\text{Mass of copper (II) oxide}=20.20g[/tex]
Hence, the theoretical yield of copper (II) oxide is 20.20 grams
