A mixture of CO₂ and Kr weighs 35.0 g and exerts a pressure of 0.708 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 0.250 atm. How many grams of CO₂ were originally present? How many grams of Kr can you recover?

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pragyavermav1 pragyavermav1 · Answer 1 · 2022-08-26T11:55:27+02:00

17.15 g of CO2 is originally present, and 17.85 g of Kr is recovered.

Since it is given that carbon dioxide is completely removed by absorption with NaOH. And, the pressure inside the container is 0.250 atm.

For Kr = 0.250 atm and pressure CO2 will be calculated as follows.

CO₂ = (0.708-0.250) atm

= 0.458 atm

Now, we will calculate the mole fraction as follows.

CO₂ = 0.458/ 0.708

= 0.646

Kr= 0.250/0.708

= 0.353

Now, we will convert it into a gram fraction as follows.

CO₂ = 0.646 x 44

= 28.424

Kr =0.353 x 83.78

= 29.57

Therefore, the total mass is calculated as follows.

Total mass = (28.424 +29.57)

= 57.994

Hence, the percentage of CO2 and Kr is calculated as follows.

CO₂ = 29/58 x 100

= 49%

Kr = 29.57/58 x 100

= 51%

Hence, the amount of CO2 and Kr present in the mixture is as follows.

CO2 in mixture 35 x 0.49

= 17.15 g

Kr = 35 x 0.51

= 17.85 g

Thus, we can conclude that 17.15 g of CO2 is originally present, and 17.85 g of Kr is recovered.

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