Amixture of xenon and helium gases at a total pressure of 927 mm hg contains xenon at a partial pressure of 719 mm hg. if the gas mixture contains 24.9 grams of xenon, how many grams of helium are present?

0,220g of He.

Explanation:

24,9g of Xe are:

24,9g Xe × ( 1mol / 131,293g ) = 0,190 moles of Xe

If total pressure is 927 mmHg and partial pressure of Xe is 719 mmHg, artial pressure of helium is:

927 mmHg - 719 mmHg = 208 mmHg

Now, Avogadro's law states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.

If temperature and volume remains constants it is possible assume the relationship that 719 mmHg are 0,190 moles to obtain moles of Helium that are:

208 mmHg × ( 0,190 mol / 719 mmHg ) = 0,0550 moles of He

In grams:

0,0550 moles of He  × ( 4,00g / 1 mol He ) = 0,220g of He

I hope it helps!

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