Question

The equation for the reaction of propyne (C3H4) with bromine (Br2) to give tetrabromopropane (C3H4Br4) is given below C3H4 + 2 Br2 → C3H4Br4 In a reaction, 4.00 g of propyne are mixed with 31.0 g of bromine. a) Which is the limiting reagent, propyne or bromine? Show all calculations. b) What is the theoretical yield of tetrabromopropane in grams? Show all calculations. c) If 13.1 g of tetrabromopropane is produced, what is the percent yield of this reaction? Show all calculations. d) If the density of liquid bromine (Br2) is 3.10 g/mL, give the volume of Br2 in mL’s to give 0.25 mols of Br2 ? [

Answer 1

The reaction between propyne (C₃H₄) and bromine (Br₂) is given as

C₃H₄ + 2 Br₂ -------> C₃H₄Br₄

As per the balanced stoichiometric equation,

1 mole C₃H₄ = 2 moles Br₂ = 1 mole C₃H₄Br₄

Molar mass of C₃H₄ = (3*12.01 + 4*1.008) g/mol = 40.062 g/mol;

Molar mass of Br₂ = (2*79.904) g/mol = 159.808 g/mol;

Molar mass of C₃H₄Br₄ = (3*12.01 + 4*1.008 + 4*79.904) g/mol = 359.678 g/mol.

a) We have 4.00 g C₃H₄ mixed with 31.0 g Br₂.

4.00 g C₃H₄ = (4.00 g C₃H₄)*(1 mole C₃H₄/40.062 g C₃H₄)*(2 mole Br₂/1 mole C₃H₄)*(159.808 g Br₂/1 mole Br₂) = 31.912 g Br₂.

31.0 g Br₂ = (31.0 g Br₂)*(1 mole Br₂/159.808 g Br₂)*(1 mole C₃H₄/2 mole Br₂)*(40.062 g C₃H₄/1 mole C₃H₄) = 3.8857 g C₃H₄.

Since we have 4.00 g C₃H₄ and 31.0 g Br₂ and the calculations show that we require 31.912 g Br₂ to completely react with 4.00 g C₃H₄, hence, Br₂ is the limiting reactant (ans).

b) The mass of tetrabromopropane, C₃H₄Br₄ is calculated from the mass of the limiting reactant.

Theoretical yield of C₃H₄Br₄ = (31.0 g Br₂)*(1mole Br₂/159.808 g Br₂)*(1 mole C₃H₄Br₄/2 mole Br₂)*(359.678 g C₃H₄Br₄/1 mole C₃H₄Br₄) = 34.8857 g ≈ 34.88 g C₃H₄Br₄ (ans).

c) The actual yield of C₃H₄Br₄ is 13.1 g.

Percent yield of C₃H₄Br₄ = (actual yield)/(theoretical yield)*100 = (13.1 g)/(34.88 g)*100 = 37.5573% ≈ 37.56% (ans).

d) Molar mass of Br₂ = 159.808 g/mol, as stated above.

Mass of 0.25 mole Br₂ = (number of mole)*(molar mass) = (0.25 mole)*(159.808 g/mol) = 39.952 g.

The density of liquid Br₂ is 3.10 g/mL; therefore, the volume of liquid Br₂ required = (mass of Br₂)/(density of Br₂) = (39.952 g)/(3.10 g/mL) = 12.8877 mL ≈ 12.90 mL (ans).