Question

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?

Answer 1

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

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

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:

$p=x\times P$

where,

$p$ = partial pressure

$P$ = total pressure = 0.751 atm

$x$ = mole fraction

For $CO_2$

$x_{CO_2}=\frac{p_{CO_2}}{P}=\frac{0.52}{0.751}=0.70$

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

For $Kr$

$x_{Kr}=\frac{p_{Kr}}{P}=\frac{0.231}{0.751}=0.30$

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

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

Percentage of $CO_2=\frac{30.8}{56}\times 100=55\%$

a) Thus Mass of $CO_2$ in mixture =$\frac{55}{100}\times 47.9=26.3g$

Percentage of $Kr=\frac{25.2}{56}\times 100=45\%$

b) Thus Mass of $Kr$ in mixture that can be recovered=$\frac{45}{100}\times 47.9=21.5g$