How to get the Theoretical Yield?

Question

if the reaction of propane + bromine --> 2,2-dibromopropane + 2 HBr

If you have 5.0 g of propane and 10.0 mL of Bromine. I balanced it but can't remember how to get the theoretical yield. the balanced equation is

CH3-CH2-CH3 + 2 Br2 --> C3H6Br2 + 2HBr

Answer 1

To know the limitig reactant we must know the moles of propane ( 5.0 g/ 44.097 g/mol =0.113) and the moles of Br2.

If we know the molarity ( or the density or the % by mass ) of the solution of Br2 we can get the moles.

Otherwise we can not answer.

If Br2 is in excess Moles 2-2, dibromopropane = 0.113

To get the mass multiply by molar mass

Answer 2

Density of bromine = 3.1028  g·cm−3

Therefore 10ml = 31.028g Br2

From your balanced equation:

1mol Prop reacts with 2 mol Br2 to produce 1mol 2,2DBP + 2mol HBr

Molar mass Prop = 44.10g

Molar mass Br2 = 159.808g

Molar mass 2,2 DBP = 201.89 g

Molar mass HBr = 80.912g

Check for limiting reactant:

44.1g prop reacts with 2*159.808 = 319.616g Br2 to produce 201.89g 2,2BDP + 161.824g HBr  Call this "mass equation"

31.028g Br2 reacts with 44.10/319.61*31.02 = 4.280g propane

Propane is in excess, Br2 is the limiting reactant.

Substitute these values into mass equation:

4.280g propane reacts with 31.028g Br2 to produce  201.89/44.10*4.28= 19.59g 2,2DBP

161.824/44.1*4.28 = 15.705g HBr

4.28g Prop + 31.028g Br2 → 19.59g 2,2DBP + 15.705g HBr

Theoretical yield of 2,2 DBP = 19.59g.