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Wate asked in Science & MathematicsChemistry · 3 months ago

CHEMISTRY HELPP!!?

A 226.4-L cylinder contains 65.5% he(g) and 34.5% kr(g) by mass at 27.0°c and 1.40 ATM total pressure. what is the mass of he in this container? The answer should be 50.3 g but Idk how to get there

THANK YOU!

2 Answers

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  • 3 months ago
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    n = PV / RT = (1.40 atm) x (226.4 L) /

    ((0.082057366 L atm/K mol) x (27.0 + 273.15) K) = 12.869 mol total

    Take a hypothetical sample of exactly 100 grams of the gaseous mixture:

    (65.5 g He) / (4.002602 g He/mol) = 16.364355 mol He

    (34.5 g Kr) / (83.798 g Kr/mol) = 0.411704 mol Kr

    (16.364355 mol He) / (16.364355 mol + 0.411704 mol) =

    0.975459 (the mole fraction of He)

    (12.869 mol total) x (0.975459) x (4.002602 g He/mol) = 50.2 g He

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  • 3 months ago

    First calculate total mol of gas in cylinder:

     PV =- nRT

    1.40 * 226.4 = n * 0.082057*300

    n = 1.40 * 226.4 / ( 0.082057*300)

    n = 12.875 total mol

    Now take 100 g of the gas mixture:

     Mass of He = 65.5 g and mass of Kr = 34.5g

    Molar mass He = 4.00 g/mol

     Mol He = 65.5g / 4.00 g/mol = 16.375 mol

    Molar mass Kr = 83.80 g/mol

     mol Kr = 34.5 g / 83.80g/mol = 0.412 mol

     Total mol in 100 g gas mixture = 16.375 + 0.412 = 16.787 mol

     Total mass gas mixture in cylinder = 12.875 mol / 16.787 mol * 100 g = 76.696 g

    Mass of He = 65.5 /100 * 76.696g = 50.24 g

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