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Chemistry Question?
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)
1 Answer
- Roger the MoleLv 73 years agoFavorite Answer
Take a hypothetical sample of exactly 100 grams 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) = 0.959011 mol Cl2
(1.58575 mol Ne) / (1.58575 mol + 0.959011 mol) = 0.623143 [the mole fraction of Ne in the mixture]
n = PV / RT = (1.07 atm) x (232 L) / ((0.08205746 L atm/K mol) x (20 + 273) K) = 10.3249 mol total
(10.3249 mol total) x (0.623143) x (20.1797 g Ne/mol) = 129.8 g Ne
