The volume of fluorine gas required to react with 2.67 g of calcium bromide to form calcium fluoride and bromine at 41.0°c and 4.31 atm is ml.

Balance Equation is as followm

                                       F₂  +  CaBr₂   →   CaF₂  +  Br₂

According to equation 1 mole of F₂ reacts with 1 mole of CaBr₂. So,
As,
                           199.89 g CaBr₂ reacts with   =   1 mole of F₂
Then,
                       2.67 g of CaBr₂ will react with   =   X mole of F₂
Solving for X,
                                        X  =  (2.67 g × 1 mole) ÷ 199.89 g

                                        X  = 0.0133 moles of F₂

Now, Covert moles of F₂ into Volume, Assuming it as Ideal gas,
So, Acc. to Ideal Gas Equation

                                          P V  = n R T
Solving for V,
                                             V  =  n R T / P

Putting Values,

               V  =  (0.0133 mol × 0.0820 atm.L.mol⁻¹.K⁻¹ × 313 K) ÷ 4.31 atm

               V  =  0.0792 L

Converting into mL,
               
                   =  0.0792 × 1000

                   =  79.2 mL