Midterm Review Q2

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tmehrazar
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Joined: Fri Apr 06, 2018 11:04 am

Midterm Review Q2

Postby tmehrazar » Sat May 05, 2018 8:25 am

2. Ammonia (NH3) reacts with oxygen (O2) to form air pollutant nitrogen oxide (NO) and water. Determine the theoretical yield of NO if 21.1 g NH3 is reacted with 42.2 g O2.

I missed the first part of the review session, is the answer to this 7.9g NO?

Sunjum Singh 1I
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Joined: Fri Apr 06, 2018 11:05 am

Re: Midterm Review Q2

Postby Sunjum Singh 1I » Sat May 05, 2018 2:06 pm

So to solve this problem you first have to write out the equation given in the problem then balance it, getting 4NH3+5O2=>4NO+6H2O. Then you take the grams of NH3 and O2 given and find the moles, multiply it by the mole ratio in the equation to get mols NO, then convert to grams. You find that O2 yields 31.7 g NO and NH3 yields 37.2 g NO, so O2 is the limiting reactant and 31.7 g NO is the theoretical yield.

Anjali_Kumar1F
Posts: 62
Joined: Fri Sep 28, 2018 12:25 am

Re: Midterm Review Q2

Postby Anjali_Kumar1F » Tue Oct 30, 2018 8:15 pm

How did you identify O2 is limiting reactant is it because 31.7 is less than 37.2. The smaller number is limiting reactant?

Anjali_Kumar1F
Posts: 62
Joined: Fri Sep 28, 2018 12:25 am

Re: Midterm Review Q2

Postby Anjali_Kumar1F » Tue Oct 30, 2018 10:36 pm

How would I solve this?


Potassium permanganate, KMnO4, is an inorganic chemical compound used for cleaning wounds. 5.00 g of KMnO4 is dissolved in a 150.00 mL flask of water. If 20.00 mL of this solution is removed and placed in a new 2nd 250.00 mL flask and filled with water, what is the concentration of the solution in the 2nd flask?

yaosamantha4F
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Joined: Fri Sep 28, 2018 12:29 am

Re: Midterm Review Q2

Postby yaosamantha4F » Sat Nov 03, 2018 1:44 am

First, calculate the molar mass of KMnO4 (158.09g/mol). Next, concert 5g KMnO4 to moles and 150mL to L and use the formula M=mol/L to calculate the molarity of the first solution. Lastly, use the equation M1V1=M2V2 equation to calculate the concentration of the second flask by using the value you previously calculated for the first solution’s molarity and the volumes given by the problem.

DavidEcheverri3J
Posts: 36
Joined: Fri Sep 28, 2018 12:26 am

Re: Midterm Review Q2

Postby DavidEcheverri3J » Sun Nov 04, 2018 4:24 pm

Sunjum Singh 1I wrote:So to solve this problem you first have to write out the equation given in the problem then balance it, getting 4NH3+5O2=>4NO+6H2O. Then you take the grams of NH3 and O2 given and find the moles, multiply it by the mole ratio in the equation to get mols NO, then convert to grams. You find that O2 yields 31.7 g NO and NH3 yields 37.2 g NO, so O2 is the limiting reactant and 31.7 g NO is the theoretical yield.

Wouldn't NH3 be the limiting reactant? since 21.1g if NH3 is 1.24 moles and 42.2g of O2 is 1.32 moles.

DavidEcheverri3J
Posts: 36
Joined: Fri Sep 28, 2018 12:26 am

Re: Midterm Review Q2

Postby DavidEcheverri3J » Sun Nov 04, 2018 4:37 pm

DavidEcheverri3J wrote:
Sunjum Singh 1I wrote:So to solve this problem you first have to write out the equation given in the problem then balance it, getting 4NH3+5O2=>4NO+6H2O. Then you take the grams of NH3 and O2 given and find the moles, multiply it by the mole ratio in the equation to get mols NO, then convert to grams. You find that O2 yields 31.7 g NO and NH3 yields 37.2 g NO, so O2 is the limiting reactant and 31.7 g NO is the theoretical yield.

Wouldn't NH3 be the limiting reactant? since 21.1g if NH3 is 1.24 moles and 42.2g of O2 is 1.32 moles.

Ohhhhh. now I get it. I was not looking at the molar ration between NH3 and O2.

Nina Do 4L
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

Re: Midterm Review Q2

Postby Nina Do 4L » Mon Nov 05, 2018 10:12 pm

Start off with what you have and convert it to grams of NO using the chemical equation.

21.1 g NH3x(1 mol NH3/ 17.031 g NH3)x(4 mol NO/4 mol NH3) [because we are trying to get grams of NO, use the chemical equation ratio] x(30.006 g NO/1mol NO)

That will give you 37.2 grams of NO produced using 21.1 grams of NH3.

Do the same for the O2.

42.2 g O2x(1 mol O2/32 g O2)x(4 mol NO/5 mol O2) {same thing, how many moles of NO is produced from O2, according to our chemical equation} x (30.006 g NO/1 mol NO)

That will give you 31.7 grams NO. Therefore, 42.2 grams of O2 will give you less product than 21.1 grams of NH3, meaning that it will run out faster and is your limiting reactant which is also your theoretical yield. Hope this helped!

brennayoung
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Joined: Fri Sep 20, 2019 12:17 am

Re: Midterm Review Q2

Postby brennayoung » Tue Nov 05, 2019 5:53 pm

yaosamantha4F wrote:First, calculate the molar mass of KMnO4 (158.09g/mol). Next, concert 5g KMnO4 to moles and 150mL to L and use the formula M=mol/L to calculate the molarity of the first solution. Lastly, use the equation M1V1=M2V2 equation to calculate the concentration of the second flask by using the value you previously calculated for the first solution’s molarity and the volumes given by the problem.

Can you explain how this works a bit further please?


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