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A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank

Answer :

Answer:

mole fraction of N_2 O = 0.330

mole of fraction SF_4 = 0.669

PRESSURE OF N_2 O = 39127.053 Pa

pressure of SF_4 = 792126.36

Total pressure   = 118253.413 Pa

Explanation:

Given data:

volume of tank 8 L

Weight of dinitrogen difluoride gas 5.53 g

weight of sulphur hexafluoride gas 17.3 g

Amount of [tex]N_2 O = \frac{5.53}{14*2 + 16} = 0.1256 mol[/tex]

amount of [tex]SF_4 = \frac{17.3}{32.1 + 19*4} = 0.254 mol[/tex]

mole fraction of [tex]N_2 O = \frac{0.1256}{0.1256 + 0.254} = 0.330[/tex]

mole of fraction[tex] SF_4 = \frac{0.254}{0.1256 + 0.254} = 0.669[/tex]

PV = nRT

P of N_2 O [tex]= \frac{0.1256 *8.31 (273 + 26.9}{0.008} = 39127.053 Pa[/tex]

mole of SF_4[tex]=\frac{0.254 *8.31*(273+26.9)}{.008} = 79126.36 Pa[/tex]

Total pressure  = 39127.053 + 79126.36 = 118253.413 Pa

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