Question

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?

Answer 1

Explanation:

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

For Kr = 0.250 atm and pressure $CO_{2}$ will be calculated as follows.

           $CO_{2}$ = (0.708 - 0.250) atm

                       = 0.458 atm

Now, we will calculate the mole fraction as follows.

            $CO_{2} = \frac{0.458}{0.708}$

                       = 0.646

               Kr = $\frac{0.250}{0.708}$

                    = 0.353

Now, we will convert into gram fraction as follows.

              $CO_{2} = 0.646 \times 44$

                          = 28.424

                   Kr = $0.353 \times 83.78$

                        = 29.57

Therefore, total mass is calculated as follows.

           Total mass = (28.424 + 29.57)

                              = 57.994

Hence, the percentage of $CO_{2}$ and Kr are calculated as follows.

          $CO_{2} = \frac{28.424}{57.99} \times 100$

                     = 49%

               Kr = $\frac{29.57}{57.99} \times 100$

                    = 51%

Hence, amount of $CO_{2}$ and Kr present i mixture is as follows.  

         $CO_{2}$ in mixture = $35 \times 0.49$

                             = 17.15 g

                  Kr = $35 \times 0.51$

                       = 17.85 g

Thus, we can conclude that 17.15 g of $CO_{2}$ is originally present and 17.85 g of Kr is recovered.