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. [4 pts]

b) What is the theoretical yield of tetrabromopropane in grams? Show all calculations. [3 pts]

c) If 13.1 g of tetrabromopropane is produced, what is the percent yield of this reaction? Show all calculations. [2 pts]

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 ? [4pts]

Answer 1

a) The no. of moles of a substance = mass of substance/molar mass of the substance

According to the given data, no. of moles of propyne = 4/40, i.e. 0.1 mol (since molar mass of propyne = 40 g mol^-1)

The no. of moles of bromine = 31/160, i.e. 0.194 mol

According to the given reaction, 1 mole of propyne requires 2 moles of bromine for the complete reaction.

i.e. 0.1 moles of propyne requires 0.2 moles of bromine, but here 0.194 (<0.2) mmol of bromine is reacted with 0.1 moles of propyne, i.e. 0.1 - (0.194/2) = 0.003 mol of propyne will remain unreacted.

Hence, bromine is the limiting reagent.

b) The theoretical yield of tetrabromopropane = (0.194/2)mol * 200 g mol^-1, i.e. 19.4 g.

c) If 13.1 g of tetrabromopropane is produced, then the percent yield of the reaction = (13.1/19.4)*100, i.e. 67.5%

d) The density of liquid bromine = 3.1 g/mL

The amount of Br₂ in its 0.25 moles = 0.25 mol * 160 g mol^-1, i.e. 40 g

= 40 g / (3.1 g/mL), i.e. 12.9 mL