Question

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 1

Mole fraction of N₂O = 0.330

Mole fraction of SF₄ = 0.669

Pressure (N₂O) = 39.12 kPa

Pressure (SF₄) = 79.12 Pa

Total Pressure = 118.25 kPa

Step-by-step explanation:

Given

Given mass of dinitrogen difluoride N₂0 =5.53 g

Given mass of sulphur hexafluoride =17.3 g

volume of tank V= 8 L= 0.008 ml

R ideal gas constant = 8.31 J / mol·K

Temperature = 26.9 C = 299.9 k

Pressure = ?

Number of moles = ?

Number of moles of N₂O in tank = given mass /molar mass =5.53 / (14.0 x 2 + 16.0) = 0.1256 mol

Number of moles SF₄ in tank = given mass /molar mass 17.3 / (32.1 + 19.0 x 4) = 0.254 mol

Mole fraction of N₂O = Number of moles of N₂0/total number of moles

= 0.1256 / (0.1256+0.254) = 0.330

Mole fraction of SF₄ = 0.254 / (0.1256+0.254= 0.669

For Pressure

PV=nRT

Pressure (N₂O) = (0.1256)(8.31)(299.9) / (0.008) = 187.22 = 39127.05 Pa = 39.12 kPa

Pressure (SF₄) = (0.254)(8.31)(299.9) / (0.008) = 79126.36 Pa= 79.12 Pa

Total Pressure = 39.12 + 79.12 = 118.25 kPa