Respuesta :
Answer: a) . Ammonia is the limiting reagent
b. Oxygen is left over and 0.1375 g of oxygen is left over.
c. The theoretical yield of [tex]NO[/tex] is 35.29 g.
d. The theoretical yield of [tex]H_2O[/tex] is 31.74 g.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
For [tex]NH_3[/tex]
Given mass of ammonia = 20.0 g
Molar mass of ammonia = 17.031 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of ammonia}=\frac{20.0g}{17.031g/mol}=1.17mol[/tex]
For [tex]O_2[/tex]
Given mass of oxygen gas = 50.0 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas}=\frac{50.0g}{32g/mol}=1.6mol[/tex]
The chemical equation for the reaction is
[tex]4NH_3+5O_2\rightarrow 4NO+6H_2O[/tex]
By Stoichiometry of the reaction:
4 moles of ammonia reacts with = 5 moles of oxygen
So 1.17 moles of ammonia will react with = [tex]\frac{5}{4}\times 1.17=1.4625mol[/tex] of oxygen
As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent and (1.6-1.4625)= 0.1375 g of oxygen is left unreacted.
Thus ammonia is considered as a limiting reagent because it limits the formation of product.
1. By Stoichiometry of the reaction:
4 moles of ammonia produces = 4 moles of [tex]NO[/tex]
1.17 moles of ammonia will produce = [tex]\frac{4}{4}\times 1.17=1.17moles[/tex] of [tex]NO[/tex]
Mass of [tex]NO=moles\times {\text{Molar Mass}}=1.17\times 30=35.29g[/tex]
Thus Theoretical yield of [tex]NO[/tex] is 35.29 grams.
2. By Stoichiometry of the reaction:
4 moles of ammonia produces = 6 moles of [tex]H_2O[/tex]
1.2 moles of ammonia will produce = [tex]\frac{6}{4}\times 1.2=1.8moles[/tex] of [tex]H_2O[/tex]
Mass of [tex]H_2O=moles\times {\text{Molar Mass}}=1.8\times 18.015=31.74g[/tex] [tex]H_2O[/tex]
Thus Theoretical yield of [tex]H_2O[/tex] is 31.74 grams.
