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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: elron on September 19, 2018, 03:21:34 PM

Title: Theoretical Yield
Post by: elron on September 19, 2018, 03:21:34 PM
ok for whatever reason this question on my study guide is killin me.  I think it's mainly because I'm so burnt out from studying but if anyone could help me explain how to get the answer I will do anything(pm me your paypal or venmo for a donation, i really appreciate your time). 

7.0 g nitrogen monoxide (NO) is reacted with 12.0 g fluorine gas (F2) to produce nitrogen fluoride and oxygen 2NO + 3F2  :rarrow: 2NF3 + O2.  What is the percent yield if 11.0g O2 is experimentally produced?

note: the answer according to the study guide answer key is 30.6% but what I really need to understand is how to get to that answer.  Thanks!
Title: Re: Theoretical Yield
Post by: chenbeier on September 19, 2018, 03:34:43 PM
You have to calculate the mole of the NO . You  can see that 2 mole NO correspond to 1 mole O2.

Then you compare the mole of 7 g NO how much it correspond to the mole of oxygen. But the given number never can obtain 11 g oxygen.

The same thing you can do by using the flourine.  3 mole F2 to 1 mole O2

In both cases 11 g is not possible.
Title: Re: Theoretical Yield
Post by: Enthalpy on September 20, 2018, 07:37:55 AM
I don't see neither how 7g NO should produce 11g O2.
Title: Re: Theoretical Yield
Post by: Traumatic Acid on October 02, 2018, 10:42:58 PM
Quote from: Enthalpy on September 20, 2018, 07:37:55 AM
I don't see neither how 7g NO should produce 11g O2.

Nope, in fact 7 grams of NO equates to 0.23 mols. This means that the most O that can be produced is 0.23 mols.

But in your equation it states 2 mols of NO, which yeilds 2 mols of O, or one mole of O2.
One mole of O2 has a mass of 32 grams. The fact that you only got 11 grams of O2 and not 32 indicates a loss. So calculate the percent yeild of O2

Actual yeild / theoretical yeild
= 11/32 = 0.34 x 100 = 34%
Not the exact same answer so I suspect the teacher was using more precise molar masses.
I'm not sure what's with the 7 grams of No, when it states in the equation that 2 moles were used. That's just sloppy stoichiometry  ;D
Title: Re: Theoretical Yield
Post by: Borek on October 03, 2018, 02:58:22 AM
Quote from: Traumatic Acid on October 02, 2018, 10:42:58 PM
Not the exact same answer so I suspect the teacher was using more precise molar masses.
I'm not sure what's with the 7 grams of No, when it states in the equation that 2 moles were used. That's just sloppy stoichiometry  ;D

No, that is not a sloppy stoichiometry, that's just a sign you don't understand the difference between balanced reaction and stoichiometry of a particular process. High school error. Not kind of a mistake I would expect from someone stating

Quote from: Traumatic Acid on October 02, 2018, 01:39:19 AM
I'm currently a second year Bachelor of Science student majoring in Chemistry and minoring in Biochemistry.