We mix 112 grams of oxygen gas with 118 grams of argon gas in a volume of 1752 mL at 74 degrees celsius. What will be the final pressure of the gas mixture?
First, I converted the grams into moles.
112g O2 * (1mol O2/ 32g O2) = 3.5 moles O2
118g Ar * (1mol Ar/ 40g Ar) = 2.95 moles O2
Then, I added the moles to get 6.45 moles ArO2
I used, P=nrT/V
=(6.45mol)(0.08206Latm/mol)(74/273)/1.752L
=104.83 atm
Is this right? Somehow, I am not getting correct sigfigs.
1 answer
I think you have a typo in your post. T is 74+ 273 and you show it as 74/273; however, you must have used 347 because my answer comes out 104.83 atm also. I would round that to 105. You have 3 s.f. in the 112 and 118. Is that 74.0 C?