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Mitra
Mitra asked in Science & MathematicsChemistry · 3 years ago

Chemistry?

Update:

A 232 L container holds 32 % Ne (g) with the remainder being Cl2 (g). The percentages given are on a per mass basis. The temperature is 20 °C and the total pressure is 1.07 atm. You may assume both gases behave ideally.

What is the mass of Ne in this container (in g)

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  • Roger the Mole
    Lv 7
    3 years ago

    n = PV / RT = (1.07 atm) x (232 L) / ((0.08205746 L atm/K mol) x (20 + 273) K) = 10.3249 mol total

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

    (32 g Ne) / (20.1797 g Ne/mol) = 1.58575 mol Ne

    (100 g - 32 g) Cl2 / (70.9064 g Cl2/mol) = 0.95901 mol Cl2

    So the mole fraction of Ne is:

    (1.58575 mol Ne) / (1.58575 mol + 0.95901 mol) = 0.62314

    (10.3249 mol total) x (0.62314) x (20.1797 g Ne/mol) = 130. g Ne

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