
In the equation- $4N{H_2} + 5{O_2} \to 4NO + 6{H_2}O$ ,$310g$ of ${O_2}$ will react with 175g of $N{H_3}$.What is the theoretical yield of $NO$,and if 197g of NO are produced, what is the percentage yield?
Answer
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Hint: To find the theoretical yield of the given compound first you know how to balance the chemical equation and you must know about the basic concept of molar mass.
Here we are going to see how to find the theoretical yield for the given equation.
Basic steps are:
Balance the chemical equation and divide the number of moles of each reactant in the balanced chemical formula to find a limiting reactant.
You can just divide the actual yield by the theoretical yield and multiply it to \[\;100\]to calculate the percentage yield.
Formula used-
$\text{Number of moles}=\dfrac{\text{Mass of substance}}{\text{Molecular Mass}}$
$ \text{Percentage Yield}$ = $\dfrac{\text{Actual Yield}}{\text{Theoritical Yield}}$ $\times {\text{100}} $
Complete answer:
We know that, Molecular masses of the molecules are:
$MrN{H_3} = 17$
$Mr{O_2} = 32 $
$MrNO = 26 $
$Mr{H_2}O = 18 $
Using the formula\[\]:
$\text{Number of moles}=\dfrac{\text{Mass of substance}}{\text{Molecular Mass}}$
Equation:
Number of Moles of $N{H_{{3_{}}}}$:
$_{} = \dfrac {175}{17} = 10.3$
Number of Moles of ${O_2}$=
$\dfrac{310}{32} = 9.69$
${O_2}$ is the limiting reagent because the least moles of this are used in the reaction.
From the balanced equation, we can able to see that the ${O_2}$ and $NO$ are in a $5:4$ ratio
$\dfrac{5}{4} = 0.8$
$9.69 \times 0.8 = 7.75$
$7.75$ moles of NO are the maximum so it can be produced from $310g$ of ${O_{_2}}$ . can use the same equation as in the first to convert this to mass step so,
$7.75 \times 26$ = $202g$
$202g$ is the theoretical yield of $NO$.
Now,
$ \text{Percentage Yield}$ = $\dfrac{\text{Actual Yield}}{\text{Theoritical Yield}}$ $\times {\text{100}} $
$\dfrac{197}{202}$ $\times 100$ = $98\% $
Theoretical yield = $202g$
Percentage yield = $98\% $
Limiting reactant
Explanation: the reactant that is used up completely is the limiting reagent. No further products are made and stop the reaction. In the given equation of the balanced chemical that describes the reaction, there are several ways to identify the limiting reagent.
Since it stops the reaction, the limiting reactant is very important...it controls the amount of product made.
The reaction goes to completion When there is no limiting reactant in a chemical equation. All of the reactants are used. Also, there is no excess.
REDUCING AGENT: The reducing agent of the given equation is ammonia. Reducing agent is the one that reduces other substances(by removal of oxygen or addition of hydrogen) and itself gets oxidized.
In this reaction, as it is being oxidized by oxygen $({O_3})$ form nitric oxide $(NO)$and water $({H_2}O)$ammonia$(N{H_3})$ is the reducing agent.
Note:
The reactant that gets consumed first in a chemical reaction is the limiting reactant or limiting reagent and therefore limits how much product can be formed. The theoretical yield is the amount of product that can be formed based on the limiting reactant.
Here we are going to see how to find the theoretical yield for the given equation.
Basic steps are:
Balance the chemical equation and divide the number of moles of each reactant in the balanced chemical formula to find a limiting reactant.
You can just divide the actual yield by the theoretical yield and multiply it to \[\;100\]to calculate the percentage yield.
Formula used-
$\text{Number of moles}=\dfrac{\text{Mass of substance}}{\text{Molecular Mass}}$
$ \text{Percentage Yield}$ = $\dfrac{\text{Actual Yield}}{\text{Theoritical Yield}}$ $\times {\text{100}} $
Complete answer:
We know that, Molecular masses of the molecules are:
$MrN{H_3} = 17$
$Mr{O_2} = 32 $
$MrNO = 26 $
$Mr{H_2}O = 18 $
Using the formula\[\]:
$\text{Number of moles}=\dfrac{\text{Mass of substance}}{\text{Molecular Mass}}$
Equation:
Number of Moles of $N{H_{{3_{}}}}$:
$_{} = \dfrac {175}{17} = 10.3$
Number of Moles of ${O_2}$=
$\dfrac{310}{32} = 9.69$
${O_2}$ is the limiting reagent because the least moles of this are used in the reaction.
From the balanced equation, we can able to see that the ${O_2}$ and $NO$ are in a $5:4$ ratio
$\dfrac{5}{4} = 0.8$
$9.69 \times 0.8 = 7.75$
$7.75$ moles of NO are the maximum so it can be produced from $310g$ of ${O_{_2}}$ . can use the same equation as in the first to convert this to mass step so,
$7.75 \times 26$ = $202g$
$202g$ is the theoretical yield of $NO$.
Now,
$ \text{Percentage Yield}$ = $\dfrac{\text{Actual Yield}}{\text{Theoritical Yield}}$ $\times {\text{100}} $
$\dfrac{197}{202}$ $\times 100$ = $98\% $
Theoretical yield = $202g$
Percentage yield = $98\% $
Limiting reactant
Explanation: the reactant that is used up completely is the limiting reagent. No further products are made and stop the reaction. In the given equation of the balanced chemical that describes the reaction, there are several ways to identify the limiting reagent.
Since it stops the reaction, the limiting reactant is very important...it controls the amount of product made.
The reaction goes to completion When there is no limiting reactant in a chemical equation. All of the reactants are used. Also, there is no excess.
REDUCING AGENT: The reducing agent of the given equation is ammonia. Reducing agent is the one that reduces other substances(by removal of oxygen or addition of hydrogen) and itself gets oxidized.
In this reaction, as it is being oxidized by oxygen $({O_3})$ form nitric oxide $(NO)$and water $({H_2}O)$ammonia$(N{H_3})$ is the reducing agent.
Note:
The reactant that gets consumed first in a chemical reaction is the limiting reactant or limiting reagent and therefore limits how much product can be formed. The theoretical yield is the amount of product that can be formed based on the limiting reactant.
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