Respuesta :
Answer: The percentage yield of the given reaction is 77.33%.
Explanation:
Moles is calculated by using the formula:
[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
- Moles of Ammonia:
Given mass of ammonia = 150g
Molar mass of ammonia = 17 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of ammonia}=\frac{150g}{17g/mol}=8.82moles[/tex]
- Moles of Oxygen
Given mass of oxygen = 150g
Molar mass of oxygen = 32 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of oxygen}=\frac{150g}{32g/mol}=4.6875moles[/tex]
For the given chemical equation:
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]
By Stoichiometry,
5 moles of oxygen reacts with 4 moles of ammonia.
So, 4.6875 moles of oxygen will react with = [tex]\frac{4}{5}\times 4.6875=3.75moles[/tex] of ammonia
As, moles of ammonia required is less than the calculated moles. Hence, ammonia is present in excess and is termed as excess reagent.
Therefore, oxygen is considered as a limiting reagent because it limits the formation of products.
By Stoichiometry of the given reaction:
5 moles of oxygen gas produces 4 moles of nitric oxide
So, 4.6875 moles of oxygen gas will produce = [tex]\frac{4}{5}\times 4.6875=3.75moles[/tex] of nitric oxide
Now, to calculate the theoretical amount of nitric oxide, we use equation 1:
Molar mass of nitric oxide = 30 g/mol
[tex]3.75mol=\frac{\text{Given mass}}{30g/mol}[/tex]
Given mass of nitric oxide = 112.5 g
Now, to calculate the percentage yield, we use the formula:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield = 87 g
Theoretical yield = 112.5 g
Putting values in above equation, we get:
[tex]\%\text{ yield}=\frac{87}{112.5}\times 100=77.33\%[/tex]
Hence, the percentage yield of the given reaction is 77.33%.
