Answer to Question #188062 in General Chemistry for Kam

Question

A mixture of krypton and xenon gases at a total pressure of 874 mm Hg contains krypton at a partial pressure of 508 mm Hg. If the gas mixture contains 14.6 grams of krypton, how many grams of xenon are present?

Answer 1

Krypton +Xenon= P_total

P_total=874mmHg

Moles of Krypton gas=14.6g×

$\frac{1mol}{83.8g/mol}$

=$0.174moles$

Partial pressure of Krypton=Mole fraction ×total pressure

Mole fraction of Krypton=$\frac{874mmHg}{508mmHg}$

$=1.720moles$

But total moles=moles×mole fraction

=0.174×1.720=$0.3moles$

Total moles=0.3moles

But Partial pressure of Xenon=(874mmHg-508mmHg)=366mmHg

Mole fraction of Xenon gas=

$\frac{874mmHg}{366mmHg}$

=2.388moles

Moles of Xenon gas =(2.388×0.3)=0.716moles

But Moles of Xenon gas=mass ×$\frac{1mole Xe}{131.29g/mole}$

Moles of Xenon gas=0.716moles

Mass of Xenon gas=$\frac{0.716}{0.01}$

=$71.6g$