[netra4662]

I just received a problem set the other day and am lost on a few of the questions and was hoping I could get some advice on how to solve them. The first question is

1. An athlete utilizes 2400.0 L of air (21% oxygen, density=1.7g/L) in running a mile. What mass of Glucose would be respired in this event assuming the production of carbon dioxide had a 15.0% yield?     I do not know where to begin this question of what I need to do in order to solve it. Any help on how I should approach this question would e greatly appreciated.

The second problem I am struggling with is

2. In a car engine the spark plug's electric discharge cause nitrogen gas to react with oxygen gas to form nitrogen monoxide gas as part of exhaust. If nitrogen monoxide exits in the atmosphere for several days, 40.0% of it will further react with oxygen to form a nitrogen dioxide gas, which can then react with water to produce nitrous acid (HNO2-acid rain) and oxygen gas in a 60.0% yield reaction. What mass of this acid rain would be produced if 8.00g of atmospheric nitrogen were consumed during an auto tip?
So far I have gotten 2N2+O2=2N2O, 2N2O+O2=2N2O2 and then N2O2=HNO2 which cannot be balanced so I don't know about what I have done so far.

I am not asking for answers to these questions but would appreciate any suggestions on how to approach these problems or advice on what I need to do. Thanks

[Borek]

1. Start with balanced reaction equation.

2. What is formula of nitrogen dioxide?

[netra4662]

Oh, I guess I had the N2 wrong. I went back and got N+O=NO, 2NO+O2=2NO2, and I also got 4NO2+2H2O=4HNO2+O2, because i found out that I forgot about an oxygen before. Does this seem like I am on track for the second question? Any tips on what to do next? Thanks

[Borek]

Now it is simple stoichiometry.

[netra4662]

This is what I have so far, I'm not sure if it is correct or not.
N+O=NO
2NO+O2=2NO2
4NO2+2H2O=4HNO2+O2
(8.00g N)(1 mole N/14.01g NO)(1 mole NO/1 mole N)(30.01g NO/1 mole NO)=17.14g NO
17.14g x 0.4=6.86
(6.86g NO)(1 mole NO/30.01g NO)(2 moles NO2/2 moles NO)(46.01g NO2/1 mole NO2)=10.52g NO2
I do not understand what to do next because of the oxygen gas in a 60% yield reaction; I do not know what this means I have to do.

[Borek]

What is definition of percent yield? If stoichiopmetric mass of the product is 100g and there is a 60% yield - what will be real mass of the product?

You forgot 40%.

Your first equation should be N₂ + O₂ - it won't change the result, but will be more correct.

[netra4662]

I thought I did the 40% when I did the 17.14 x .4, but is that the correct way to do that? And for the percent yield, which I know is actual amount made/most amount that can be theoretically made, would I just do the rest of the problem like I have been and then just multiply my answer by 0.6?

[Borek]

I thought I did the 40% when I did the 17.14 x .4, but is that the correct way to do that?

Sorry, missed that step.

And for the percent yield, which I know is actual amount made/most amount that can be theoretically made, would I just do the rest of the problem like I have been and then just multiply my answer by 0.6?

Yep

[netra4662]

I finished the problem like this:
(10.52g NO2)(1 mole NO2/46.01g NO2)(4 moles HNO2/4 moles NO2)(47.02g HNO2/1 mole HNO2)=10.75g HNO2
10.75g HNO2 x .6=6.45g Acid rain produced.
Just wanted to know if this looks right, And with the N+O instead of N2+O2, are you sure that would not change the result? I would think that it would because the formula mass would be different in my first step of the problem. Am i right or is it alright leaving it how it is?
Also, I still do not have any idea to approach the first problem. Thanks for all of your help

[Borek]

I finished the problem like this:
(10.52g NO2)(1 mole NO2/46.01g NO2)(4 moles HNO2/4 moles NO2)(47.02g HNO2/1 mole HNO2)=10.75g HNO2
10.75g HNO2 x .6=6.45g Acid rain produced.

Looks OK to me, although I have just skimmed the calculations.

And with the N+O instead of N2+O2, are you sure that would not change the result? I would think that it would because the formula mass would be different in my first step of the problem. Am i right or is it alright leaving it how it is?

Changing to N₂ and so on will not change calculation result, as it doesn't change molar ratio - there is 1 molecule produced per every atom of nitrogen. But you better change to your reaction to N₂ version, as we always put diatomics as such into equations.

Also, I still do not have any idea to approach the first problem. Thanks for all of your help

As for the first problem - I would start with the balanced reaciton equation of glucose oxidation. But then I have no idea what it means by "respiring glucose". Either it is some error in the text or my English fails me. I can guess that it is really about glucose oxidation and that this oxidation is not complete, so that there is only 15% of expected CO₂ produced - but even then, this information is irrelevant to the question, as all you can do is to find mass of glucose that reacts with given amount of oxygen.

Perhaps someone else will have an idea what is going on, question doesn't make sense to me.

[netra4662]

I got a little bit of the first question figured out but again am stuck and don't know what to do with the rest of the information. This is what I have gotten so far:
The formula- C6H12O6+6O2=6CO2+6H2O
2400.0L air x .21=504L O2 x 1.17=589.68g O2
(589.68g O2)(1 mole O2/32g O2)=18.43 moles O2
I think this is correct but am not positive and don't know what to do next with the 15% CO2 and the rest of the question. And I believe that the question is asking what amount of glucose is used. Thanks

[Borek]

Amount of glucose is given by the reaction stoichiometry.

[netra4662]

Tried to figure it out but don't know if I got anywhere with it. This is what I had before:
The formula- C6H12O6+6O2=6CO2+6H2O
2400.0L air x .21=504L O2 x 1.17=589.68g O2
(589.68g O2)(1 mole O2/32g O2)=18.43 moles O2

And this is what I've gotten since then:
(589.68g O2)(1 mole O2/32.00g O2)(1 mole C6H12O6/6 moles O2)(180.18g Glucose/1 mole glucose)=553.38g glucose
(553.38g glucose)(1 mole glucose/180.18g glucose)(6 moles CO2/1 mole glucose)(18.02g CO2/1 mole CO2)=332.06g CO2

I'm not sure if what I did is going to help me solve the problem at all. I also do not understand where the 15% yield of CO2 comes in, as well as if the limiting reactant in the equation will have anything to do with it and if it does, where it will come in. Thanks again

[Borek]

All I can tell you is that you have correctly calculated amount of glucose. CO₂ molar mass is not 18.02, that is molar mass of water. No idea what to do with 15%.