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Calculating Molar Concentrations and Partial Pressures of AsF3, F2, and AsF5

This document outlines a series of calculations for the molar concentrations and partial pressures of arsenic trifluoride (AsF3), fluorine (F2), and arsenic pentafluoride (AsF5) in a gaseous reaction system. Given an initial mass of 55.8 g of AsF5, the calculations use the ideal gas law, and stoichiometric relationships to determine the concentrations and partial pressures of the species involved. The equilibrium constant (K_eq) is also derived to help understand the dynamic state of the reaction.

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Calculating Molar Concentrations and Partial Pressures of AsF3, F2, and AsF5

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  1. 2008 B 1

  2. ai mol AsF5 = (55.8 g of AsF5 )* (1 mol AsF5) (169.9 g) mol AsF5 = 0.328 mol [AsF5] = 0.328 mol AsF5 = 0.0313M 10.5L

  3. a ii. • PV=nRT • P = (0.328 mol) * (0.0821 L*atm mol-1 K-1) *(378K) (10.5L) P= 0.969 atm

  4. b. K = [AsF3 ] [F2] [AsF5] c. i. 100%- 27.7 % = 72.3% AsF5 = 0.723 * 0.0313M = 0.0226M

  5. C. ii. [AsF3] = [F2] = 0.277 * [AsF5] = 2.77 * 0.0313 M = 0.00867 M Keq = [AsF3][F2] = [0.00867] [0.00867] [AsF5] [0.0226]

  6. d. mol AsF5 = 0.0226 M *10.5 L = 0.237 mol mol AsF3 = mol F2 = 0.00867 M * 10.5L = 0.0910 mol

  7. mol fraction F2 = mol F2 mol F2 * mol AsF3 * mol AsF5 mol fraction F2 = (0.00864) (0.00864+0.00864+0.0226) mol fraction F2 = 0.217

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