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Be sure to answer all parts. A mixture of CO2 and Kr weighs 47.9 g and exerts a pressure of 0.751 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 0.231 atm. (a) How many grams of CO2 were originally present? (b) How many grams of Kr can you recover?

1 Answer

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Answer:a) Mass of
CO_2 in mixture = 26.3 grams

b) Mass of
Kr in mixture that can be recovered= 21.5 grams

Step-by-step explanation:

Given :

Total Pressure = pressure of krypton + pressure of carbon dioxide = 0.751 atm

pressure of krypton = 0.231 atm

Thus pressure of carbon dioxide = 0.751 - 0.231 =0.52 atm

As we know:


p=x* P

where,


p = partial pressure


P = total pressure = 0.751 atm


x = mole fraction

For
CO_2


x_(CO_2)=(p_(CO_2))/(P)=(0.52)/(0.751)=0.70


{\text {Mass of} CO_2}=moles* {\text {Molar mass}}=0.70* 44=30.8g

For
Kr


x_(Kr)=(p_(Kr))/(P)=(0.231)/(0.751)=0.30


{\text {Mass of} Kr}=moles* {\text {Molar mass}}=0.30* 84=25.2g

Total mass = Mass of
CO_2 + Mass of krypton = 30.8 + 25.2 = 56 g

Percentage of
CO_2=(30.8)/(56)* 100=55\%

a) Thus Mass of
CO_2 in mixture
=
(55)/(100)* 47.9=26.3g

Percentage of
Kr=(25.2)/(56)* 100=45\%

b) Thus Mass of
Kr in mixture that can be recovered
=
(45)/(100)* 47.9=21.5g

User Arpit Gupta
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