Question

10.0 grams of carbon dioxide and 20.0 grams of oxygen are mixed in a 50.0 L container at 30.0 degrees celcius. What is the pressure of this mixture in mm Hg? I defined my variables. so i had Mass=30.0g Volume=50.0L Temperature=303 Kelvin pressure=x.
then i used PV*GFM=mRT.
(x)(50.0L)(60)=30.0G(62.4)(303). The correct answer is 322 mmHG. But i'm not getting that. can anyone help? thanks:)

Answer 1

Do the O2 and the CO2 SEPARATELY.
Add the two pressures at the end to get total pressure.

Answer 2

okay thanks:) I really appreciate your help:)

Answer 3

I didn't get the right answer....

Answer 4

First CO2
10 grams CO2 is how many moles?
C = 12
O2 = 32
CO2 = 44 grams per mole
so
n of CO2 = 10/44 = .227 moles
P V = n R T
P of CO2 = .227 R ( T/V)
NOW O2
20 grams O2
32 grams / mole
so 20/32 = .625 moles O2
P of O2 = .625 R (T/V)

add P of CO2 to P of O2 to get total pressure

Answer 5

we learned that R is 62.4mmHG. so how would i do the problem using that value?

Answer 6

hmm
in chemistry units usually R = .08206 liter atmospheres / moles degree Kelvin
let me see if I can convert that to mm Hg

Answer 7

OK, multiply by 760 mm Hg / atm and I get
R = 62.37 L mmHg/mol deg K
now I will try to do it your class' way

Answer 8

you're right .0821 is atm and 8.31 is kPa and 62.4 is mmHG.

Answer 9

this is what i tried doing, (x)(50.0)(44)=(20.0)(62.4)(303) for carbon dioxide then i did the same for water but plugged in different numbers for the GFM and the grams.

Answer 10

First CO2
10 grams CO2 is how many moles?
C = 12
O2 = 32
CO2 = 44 grams per mole
so
n of CO2 = 10/44 = .227 moles
P V = n R T
P of CO2 = .227 R ( T/V)
P of CO2 = .227* 62.4 * 303/50 = 85.83
NOW O2
20 grams O2
32 grams / mole
so 20/32 = .625 moles O2
P of O2 = .625 R (T/V) =.625*62.4*303/50 = 236.34
sum = P = 236 + 86 = 322 mmHg
I agree

Answer 11

THANK YOU DAMON, YOU'RE THE MANNN!!!

Answer 12

Not water, CO2 and O2

Answer 13

huhh, what??