Question

Profiles Tab Window HelpE- CengageNOW Online texgmenttakeCovalentActivity.do?locator assignment-takeTUTOR Percent YieldCengage: Digital Course S xResubmit[References]Consider this reaction, which occurs in the atmosphere and contributes to photochemical smog:N₂(9) + O₂(g) -2NO(g)If there is 16.28 g N₂ and excess O₂ present, the reaction yields 27.5 g NO. Calculate the percent yield for thereaction.Show Approach Show Tutor StepsE-OWL2| Online teaching x E OWLA21 Assignments | D902IncorrectYou can work though the tutor steps and submit your answer again.2

Answer 1

`78.83\text{ \%}`

Step-by-step explanation:

Here, we want to calculate the percentage yield of the reaction

Mathematically, we have that as:

`\frac{Actual\text{ Yield}}{Theoretical\text{ Yiled }}\text{ }*\text{ 100 \%}`

The mass of the product NO given in the question is the actual yield which is 27.50 g

Now, let us get the theoretical yield

We find the number of moles of nitrogen gas that reacted

Mathematically, that is the mass of nitrogen gas divided by the molar mass

The molar mass of the nitrogen gas is 28 g/mol

The number of moles is thus:

`(16.28)/(28)\text{ = 0.5814 mol}`

From the equation of reaction, 1 mole of N2 yields 2 moles of NO

Thus, 0.5814 mol of N2 will yield:

`0.5814\text{ }*\text{ 2 = 1.1628 mol}`

Now, to get the mass of NO produced, we multiply this number of moles by the molar mass of NO

The molar mass of NO is 30 g/mol

The mass that was produced is thus:

`30\text{ }*\text{ 1.1628 = 34.884 g}`

Finally, we have the percentage yield as:

`(27.50)/(34.884)\text{ }*\text{ 100 \% = 78.83 \%}`