If 10.0 moles of O₂ are reacted with excess NO in the reaction below, and only 4.2 mol of NO₂ were collected, then what is the percent yield for the reaction? 2 NO (g) + O₂ (g) → 2 NO₂ (g)

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

From the question

Actual yield = 4.2 mole

∴ Percent yield for the reaction =

Percent yield for the reaction =

Percent yield for the reaction = 21%

Hence, the percent yield for the reaction is 21%

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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%

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8

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