Exercise B2.7 Gaseous chlorine trifluoride (CIF3) can be prepared by the reactio

Question

Exercise B2.7 Gaseous chlorine trifluoride (CIF3) can be prepared by the reaction of dichlorine and difluorine. Cl2(g) + 3 F2(g)- 2 CIF:G) A 5.00 L flask is filled with 10.0 g Cl2 and 10.0 g F2, and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature of the flask is 125°C. (a) Determine the partial pressure of CIF3 in the flask when the reaction is complete. (b) Determine the total pressure in the flask when the reaction is complete.

Explanation / Answer

Cl2(g) + 3F2(g) - - - - - > 2ClF3

stoichiometrically, 1mole of Cl2 react with 3mole of F2

no of mole of Cl2 = 10g/35.45g/mol = 0.28209mole

no of mole of F2 = 10g /18.998g/mol = 0.52637mole

0.28209mole of Cl2 require 0.84627mole of F2 to react completely, but available mole of F2 is 0.52637mole. So, F2 is limiting and Cl2 excess.

0.52637mole of F2 give (2/3)*0.52637 = 0.35091mole of ClF3

0.52637mole of F2 react with 0.52637/3 = 0.17546 mole of Cl2

remaining mole of Cl2 = 0.28209 - 0.17546 = 0.10663

Now come to the question

a) Ideal gas equation is

PV = nRT

P = nRT/V

P = pressure

n = no of mole, 0.35091

R= 0.082057(L atm/mol K)

T = temperature, 125°C = 398.15K

V = 5L

P = 0.35091mol*0.082057(L atm/mole K) * 398.15K/5L

= 2.293atm

Therefore,

Partial pressure of ClF3 = 2.293atm

b) total mole = no of mole of Cl2 unreacted + no of mole of ClF3 produced

= 0.10663 + 0.35091

= 0.45754

applying ideal gas equation

P = 0.45754mol*0.082057(L atm/mol K) *398.15K/5L   

= 2.990atm

Therefore,

Total pressure of in the flask = 2.990atm

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