# Can someone help me with a couple Chemistry Questions?

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#1
Hey guys, I have a couple Chem questions that I need answers to and I was hoping you guys can help me out!

Question 1:
K2O2 is used to absorb the CO2 produced by people in space vehicles according to the following equation:
2K2O2 + 2CO2 -> 2K2CO3 + O2
If a person exhales 3 L of air per minute, and CO2 comprises 3.4% of exhaled air, (at 25 degrees celsius and 728 mmHg). The amount of K2O2 needed for a 5 day trip is what?
A.) 4400 grams
B.) 3100 grams
C.) 3800 grams
D.) 2500 grams
0
3 years ago
#2
(Original post by Justin05)
Hey guys, I have a couple Chem questions that I need answers to and I was hoping you guys can help me out!

Question 1:
K2O2 is used to absorb the CO2 produced by people in space vehicles according to the following equation:
2K2O2 + 2CO2 -> 2K2CO3 + O2
If a person exhales 3 L of air per minute, and CO2 comprises 3.4% of exhaled air, (at 25 degrees celsius and 728 mmHg). The amount of K2O2 needed for a 5 day trip is what?
A.) 4400 grams
B.) 3100 grams
C.) 3800 grams
D.) 2500 grams
My best attempt is as follows:

3L m-1 x 60 = 180L Hr^-1

180L x 24 = 4320L Day^-1

4320L x 5 = 21600 L over 5 Days

21600/100*3.4 = 734.4L of CO2 over 5 days

Calculating Vm or the Volume a mole of an ideal gas occupies under certain conditions is as follows:

Pv=nRT

v = nRT/P

therefore:

RT/P = V/n or Volume per mol of gas = Vm

Taking the gas constant as 62.3637 L.mmHg.K^-1.Mol^-1

T = 298K
P = 728

(62.3637x298)/728 = 25.528 dm3 mol^-1

n = V/Vm

Therefore 734.4dm3/25.528 = 28.768 mol of CO2

1:1 ratio so 28.768 mol of K2O2 are needed

110 = Mr of K2O2 which I take as 2(39)+2(16)

28.768*(110) = 3164.48g required which isn't far off 3100g. There'd be a bit of an excess.

Do you have a mark scheme? They could well have used different values for R or even just given you Vm, but the fact that they gave you values for P and T suggested to me that they wanted you to calculate it.

For future ref: Put this in the Chemistry section.
1
3 years ago
#3
(Original post by Justin05)
x
Moved this to 'Chemistry' for you
1
#4
(Original post by Steljoy)
Moved this to 'Chemistry' for you
Oh, Thank you! Sorry, first time using this site!
1
3 years ago
#5
(Original post by Justin05)
Oh, Thank you! Sorry, first time using this site!
That's okay! Don't worry about it buddy!

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