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

Pressure of Kr = 0.193 atm

Pressure of CO₂ = 0.729 - 0.193 = 0.536 atm

Partial pressure is directly proportial to mole fraction. THus, mole fraction can be calculated as:

CO₂ = Pressure due to CO₂/ total pressure = 0.536atm/0.729atm = 0.735

similary, Kr = 0.193/ 0.729 = 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

Percent of gas in mixture is

% CO₂ = (32.34 /54.54)* 100 % = 59.30%

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

% Kr =22.20/ 54.54 * 100 % = 40.70 %

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