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A mixture 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?

Respuesta :

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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