Question

A mixture of CO2 and Kr weighs 41.0 g and exerts a pressure of 0.729 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 0.193 atm. (a) How many grams of CO2 were originally present? (b) How many grams of Kr can you recover?

Answer 1

Answer:

a) 24.31 g

b) 16.69 g

Explanation:

A mixture of CO2 and Kr weighs 41.0 g and exerts a pressure of 0.729 atm in its container.

After the CO2 is completely removed by absorption with NaOH(s), the pressure in the container is 0.193 atm.

Therefore, Pressure of Kr = 0.193 atm

Pressure of CO2  = 0.729 - 0.193 = 0.536 atm

Their mole fraction can be also determined as follows:

CO2 = $\frac{Presssure due to CO_2}{Total pressure}$

CO2 = $\frac{0.536}{0.729}$

= 0.735

Also; for Kr ; we have

Kr = $\frac{0.193}{0.729}$

Kr = 0.265

Molar mass of CO2 = 44 g/mol

Molar mass of Kr = 83.78 g/mol

Mass of CO2 = mole fraction * molar mass = 0.735 * 44 = 32.34

Mass of Kr = 0.265 *  83.78 = 22.20

Total mass = 32.34 +22.20 = 54.54

The Percentage of gas in mixture is  as follows:

% CO2 = $\frac{32.34}{54.54}$* 100 %

= 0.5930

= 59.30%  

(a) Mass of CO2 in mixture = 0.5930* 41 g = 24.31 g

% Kr =$\frac{22.20}{ 54.54}$ * 100 %

= 0.407

= 40.70 %

(b) Mass of Kr in mixture = 0.407 * 41 = 16.69 g