Question

1. Chlorine and fluorine gases can react to form chlorine trifluoride: Cl₂(g) +    3F₂(g) --> 2ClF₃(g)

What is the theoretical yield in grams of chlorine trifluoride (ClF₃)if a 2.0 L container at 350 K initially contains Cl₂ gas at a partial pressure of 1.0 atm and fluorine gas at a partial pressure of 2.0 atm?

Hint: This is a limiting reactant problem.

A.		8.6 g ClF3
B.		21.5 g ClF3
C.		13 g ClF3
D.		19.3 g ClF3

Answer 1

Ans. #Step 1: Calculate moles of Cl₂:

Given, Volume, V = 2.0 L     ; Pressure, P = 1.0 atm         ; Temperature, T = 350 K

# Using Ideal gas equation:            PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Putting the values in equation 1-

            1.00 atm x 2.0 = n x (0.0821 atm L mol^-1K^-1) x 350 K

            Or, 2.0 atm L = n x 28.735 atm L mol^-1

            Or, n = 2.0 atm L / (28.735 atm L mol^-1) = 0.0696 mol

Hence, number of moles of Cl₂ taken = 0.0696 mol

#Step 2: Calculate moles of F₂:

Given, Volume, V = 2.0 L     ; Pressure, P = 2.0 atm         ; Temperature, T = 350 K

# Using Ideal gas equation:            PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Putting the values in equation 1-

            2.00 atm x 2.0 = n x (0.0821 atm L mol^-1K^-1) x 350 K

            Or, 2.0 atm L = n x 28.735 atm L mol^-1

            Or, n = 4.0 atm L / (28.735 atm L mol^-1) = 0.0696 mol

Hence, number of moles of Cl₂ taken = 0.1392 mol

# Step 3: Determine the limiting reactant:

            Balanced Reaction:               Cl₂(g) + 3 F₂(g) -------> 2 ClF₃(g)

Theoretical molar ratio of reactants in balanced reaction = Cl₂ : F₂ = 1 : 3

Experimental molar ratio of reactants = Cl₂ : F₂ = 0.0696 : 0.1392 = 1 : 2

Compare the theoretical and experimental molar ratios of reactants, the number of moles of F₂ is less than theoretical value of 3 moles whereas that of Cl₂ is constant at 1 mol.

Hence, F₂ is the limiting reagent.

# Step 4: Calculate Theoretical yield

The formation of product follows the stoichiometry of limiting reactant.

According to the stoichiometry of balanced reaction, 3 mol F₂ forms 2 mol product, ClF₃.

So,

Theoretical moles of ClF₃ formed = (2 / 3) x moles of F₂

                                                = (2 / 3) x 0.1392 mol

                                                = 0.0928 mol

Theoretical mass of ClF₃ formed = Theoretical moles x Molar mass

                                                = 0.0928 mol x (92.4479096 g/ mol)

                                                = 8.58 g

                                                = 8.6 g

Thus, correct option is – A. 8.6 g ClF₃