Respuesta :
Answer:a) Mass of [tex]CO_2[/tex] in mixture = 26.3 grams
b) Mass of [tex]Kr[/tex] 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:
[tex]p=x\times P[/tex]
where,
[tex]p[/tex] = partial pressure
[tex]P[/tex] = total pressure = 0.751 atm
[tex]x[/tex] = mole fraction
For [tex]CO_2[/tex]
[tex]x_{CO_2}=\frac{p_{CO_2}}{P}=\frac{0.52}{0.751}=0.70[/tex]
[tex]{\text {Mass of} CO_2}=moles\times {\text {Molar mass}}=0.70\times 44=30.8g[/tex]
For [tex]Kr[/tex]
[tex]x_{Kr}=\frac{p_{Kr}}{P}=\frac{0.231}{0.751}=0.30[/tex]
[tex]{\text {Mass of} Kr}=moles\times {\text {Molar mass}}=0.30\times 84=25.2g[/tex]
Total mass = Mass of [tex]CO_2[/tex] + Mass of krypton = 30.8 + 25.2 = 56 g
Percentage of [tex]CO_2=\frac{30.8}{56}\times 100=55\%[/tex]
a) Thus Mass of [tex]CO_2[/tex] in mixture =[tex]\frac{55}{100}\times 47.9=26.3g[/tex]
Percentage of [tex]Kr=\frac{25.2}{56}\times 100=45\%[/tex]
b) Thus Mass of [tex]Kr[/tex] in mixture that can be recovered=[tex]\frac{45}{100}\times 47.9=21.5g[/tex]
