Question

A mixture of CO2 and Kr weighs 31.7 g and exerts a pressure of 0.665 atm in its container. Since Kr is expensive, you wish to recover it from the mixture. After the CO2 is completely removed by absorption with NaOH(s), the pressure in the container is 0.309 atm.
(a) How many grams of CO2 were originally present?
(b) How many grams of Kr can you recover?

Answer 1

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 = $n_1=\frac{x}{44 g/mol}=$

Moles of krypton gas = $n_2=\frac{y}{84 g/mol}=$

Mole fraction of krpton =$\chi '$

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 :

$p'=P\times \chi '$

$0.309 atm=0.665 atm\times \frac{n_2}{n_1+n_2}$

$0.309 atm=0.665 atm\times \frac{\frac{y}{84}}{\frac{x}{44}+\frac{y}{84}}$

$0.4645=\frac{\frac{y}{84}}{\frac{x}{44}+\frac{y}{84}}$..[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.