Answer:
11.94 grams of carbon dioxide were originally present.
19.94 grams of krypton can you recover.
Explanation:
Mass of carbon dioxide gas = x
Mass of krypton gas = y
x + y = 31.7 g
Moles of carbon dioxide gas = ![n_1=\frac{x}{44 g/mol}=](https://tex.z-dn.net/?f=n_1%3D%5Cfrac%7Bx%7D%7B44%20g%2Fmol%7D%3D)
Moles of krypton gas = ![n_2=\frac{y}{84 g/mol}=](https://tex.z-dn.net/?f=n_2%3D%5Cfrac%7By%7D%7B84%20g%2Fmol%7D%3D)
Mole fraction of krpton =![\chi '](https://tex.z-dn.net/?f=%5Cchi%20%27)
Total pressure of the mixture = P = 0.665 atm
Partial pressure of carbon dioxide gas = p
Partial pressure of krypton gas before removal of carbon dioxide gas = p'
Partial pressure of krypton gas after removal of carbon dioxide gas = p'' = 0.309 atm
p' = p'' = 0.309 atm
0.665 atm = p + 0.309 atm
p = 0.665 atm - 0.306 atm = 0.359 atm
Partial pressure of krypton can also be given by :
![p'=P\times \chi '](https://tex.z-dn.net/?f=p%27%3DP%5Ctimes%20%5Cchi%20%27)
![0.309 atm=0.665 atm\times \frac{n_2}{n_1+n_2}](https://tex.z-dn.net/?f=0.309%20atm%3D0.665%20atm%5Ctimes%20%5Cfrac%7Bn_2%7D%7Bn_1%2Bn_2%7D)
![0.309 atm=0.665 atm\times \frac{\frac{y}{84}}{\frac{x}{44}+\frac{y}{84}}](https://tex.z-dn.net/?f=0.309%20atm%3D0.665%20atm%5Ctimes%20%5Cfrac%7B%5Cfrac%7By%7D%7B84%7D%7D%7B%5Cfrac%7Bx%7D%7B44%7D%2B%5Cfrac%7By%7D%7B84%7D%7D)
..[2]
Solving [1] and [2]:
x = 11.94 g
y = 19.76 g
11.94 grams of carbon dioxide were originally present.
19.94 grams of krypton can you recover.