Respuesta :
Answer:
Percent yield = 21%
Explanation:
Find:
Percent yield
Computation:
1 mol O₂ produces 2 mol NO₂
So,
10 mol O₂ produces 20 mol NO₂
Percent yield = [Actual yield / Theoretical yield]100
Percent yield = [4.2 / 20]100
Percent yield = 21%
The percent yield for the reaction is 21%
From the question,
We are to determine the percent yield for the reaction.
First, we will determine the theoretical yield for the reaction
From the given balanced chemical equation for the reaction
2 NO (g) + O₂ (g) → 2 NO₂ (g)
This means,
2 moles of NO reacts with 1 mole of O₂ to produce 2 moles of NO₂
From the given information,
Number of moles of O₂ that reacted = 10.0 moles
Since,
2 moles of NO reacts with 1 mole of O₂ to produce 2 moles of NO₂
Then,
2×10 moles of NO will be required to react with the 10 moles of O₂ to produce 2×10 moles of NO₂
That is,
20 moles of NO will be required to react with the 10 moles of O₂ to produce 20 moles of NO₂
∴ 20 moles of NO₂ will be produced during the reaction.
This is the theoretical yield
Now, for the percent yield of the reaction,
Using the formula
[tex]Percent\ yield = \frac{Actual\ yield }{Theoretical\ yield} \times 100\%[/tex]
From the question
Actual yield = 4.2 mole
∴ Percent yield for the reaction = [tex]\frac{4.2}{20 }\times 100\%[/tex]
Percent yield for the reaction = [tex]\frac{420}{20}\%[/tex]
Percent yield for the reaction = 21%
Hence, the percent yield for the reaction is 21%
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