Amixture of helium and krypton gases at a total pressure of 904 mm hg contains helium at a partial pressure of 718 mm hg. if the gas mixture contains 0.777 grams of helium, how many grams of krypton are present?

If helium has partial pressure of 718 mmHg, the krypton must have (904 mmHg - 718 mmHg = ) 186 mmHg. Since temperature and volme are the same for each gas, the mole ratio of helium to krypton must be 718 / 186. The dimensional analysis then yields:

0.777 g He * 1 mol He / 4.00 g He * 186 mol Kr / 718 mol He * 83.80 g Kr / mol Kr = 4.22 g Kr.

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