Question

If the percentage yield of the given reaction is 30%, how many total moles of the gases will be produced if 8 moles of $NaN{{O}_{3}}$ are initially taken?
$4NaN{{O}_{3}}\to 2N{{a}_{2}}O+2{{N}_{2}}+5{{O}_{2}}$

Answer 1

Write the chemical reaction with states of reaction and products and then use percentage yield formula to calculate the actual yield of the gases produced in the reaction.

Formula used: We will use the percentage yield formula i.e., $\dfrac{\text{Actual yield}}{\text{Theoretical Yield}}\times 100=\text{percent yield}$

Complete step-by-step answer: As we know that percent yield tells us about how many moles of a product will actually be produced by a chemical reaction for every 100 moles of this product that could theoretically be produced by the reaction. And since we are given 30% yield, which means that for every 100 moles of a product that the reaction could produce, we will only get 30 moles.
Now the above decomposition reaction can be described as follows:-
$4NaN{{O}_{3(s)}}\xrightarrow{\Delta }2N{{a}_{2}}{{O}_{(s)}}+2{{N}_{2(g)}} \uparrow +5{{O}_{2(g)}} \uparrow$
By this representation, we can conclude that we have to only calculate the actual yield of oxygen and nitrogen gas.
-Since 4 moles of $NaN{{O}_{3}}$ give 2 moles of ${{N}_{2}}$ gas. Then 8 moles of $NaN{{O}_{3}}$will give following moles of${{N}_{2}}$ gas:-
$8\text{ moles NaN}{{\text{O}}_{3}}\times \dfrac{2\text{ moles }{{\text{N}}_{2}}}{4\text{ moles NaN}{{\text{O}}_{3}}}=4\text{ moles }{{\text{N}}_{2}}$
-Since 4 moles of $NaN{{O}_{3}}$ give 5 moles of ${{O}_{2}}$ gas. Then 8 moles of $NaN{{O}_{3}}$will give following moles of${{O}_{2}}$ gas:-
$8\text{ moles NaN}{{\text{O}}_{3}}\times \dfrac{\text{5 moles }{{\text{O}}_{2}}}{4\text{ moles NaN}{{\text{O}}_{3}}}=10\text{ moles }{{\text{O}}_{2}}$
4 moles of ${{N}_{2}}$ gas and 10 moles of ${{O}_{2}}$ gas is the theoretical yield that will be produced from the reaction.

Now, we can use the percentage yield formula to calculate the actual yield:-
$\dfrac{\text{Actual yield}}{\text{Theoretical Yield}}\times 100=\text{percent yield}$

On rearranging it, we get:-
$\dfrac{\text{Percent yield}\times \text{ Theoretical Yield}}{100}=\text{actual yield}$

-The actual yield of ${{N}_{2}}$ gas produced = $\dfrac{30\times \text{4 moles of }{{\text{N}}_{2}}}{100}=\text{1}\text{.2 moles of }{{\text{N}}_{2}}$
- The actual yield of ${{O}_{2}}$ gas produced =$\dfrac{30\times 10\text{ moles of }{{\text{O}}_{2}}}{100}=3\text{ moles of }{{\text{O}}_{2}}$
Hence, total moles of the gases produced = 1.2 + 3 = 4.2 moles

Note: We can use an alternative method by directly assuming that from 8 moles of$NaN{{O}_{3}}$, only 30% of the reactant took part in this reaction and then start with further calculations as follows:-
The 30% of $NaN{{O}_{3}}$that took part in reaction = $\dfrac{30\times 8\text{ moles of NaN}{{\text{O}}_{3}}}{100}=2.4\text{ moles of NaN}{{\text{O}}_{3}}$
Now, the only job is left is to calculate actual yield of gases from actual part of the reactant that that participated in the reaction:-
-Since 4 moles of $NaN{{O}_{3}}$ give 2 moles of ${{N}_{2}}$ gas. Then 2.4 moles of $NaN{{O}_{3}}$will give following moles of${{N}_{2}}$ gas:-
$2.4\text{ moles NaN}{{\text{O}}_{3}}\times \dfrac{2\text{ moles }{{\text{N}}_{2}}}{4\text{ moles NaN}{{\text{O}}_{3}}}=1.2\text{ moles }{{\text{N}}_{2}}$
-Since 4 moles of $NaN{{O}_{3}}$ give 5 moles of ${{O}_{2}}$ gas. Then 8 moles of $NaN{{O}_{3}}$will give following moles of${{O}_{2}}$ gas:-
$2.4\text{ moles NaN}{{\text{O}}_{3}}\times \dfrac{\text{5 moles }{{\text{O}}_{2}}}{4\text{ moles NaN}{{\text{O}}_{3}}}=3\text{ moles }{{\text{O}}_{2}}$