Question

Please find theoretical yield and limiting reagent for both reactions. Thanks!

Reactions and/or chemicals of interest: HBr Rr+ 1-hexene 1-bromohexane Br 2-bromohexane Molar Mass: .llolmo Volume used: S. Volume used: 1:0 ml Concen Concentration: _ 1.8 M Density: 0.6730mL Moles: - Boiling Point: 6.3°C Since both products have the same molar mass, calculate the yield of the mixture of both products. Moles: Molar Mass: 105.07! gimur Limiting Reagent: Theoretical Yield: X + HBr - X 3,3-dimethyl-1-butene 1-bromo-3,3-dimethylbutane 2-bromo-3,3-dimethylbutane Br 2-bromo-2,3- dimethylbutane Since all three products have the same molar mass, calculate the yield of the mixture of products. ome Molar Mass: 84.169 ima Volume used. Some Volume used: Concentration: - Density: Oos gimc Moles: Moles: 1 Boiling Point: 41.2°C Moles: Molar Mass: 165.07 gimor Limiting Reagent: Theoretical Yield:

Answer 1

Let's see stiochiometric reaction -

1-hexene. + HBr $\LARGE \rightarrow$ 1-bromobutane + 2 -bromobutane

1) 1-hexene

# Molar mass of 1-hexene= 84.16 g/mol

# Amount of 1-hexene = 1 mL

# density of 1-hexene= 0.65 g/mL

# we know that , mass = density x volume

= 0.65 g/mL x 1.0 mL

= 0.65 g of 1-hexene.

Here key raw material is 1-hexene , so moles of 1-hexene decides mass of theoretical product , hence it is treated as limiting reagent .

Moles of 1-hexene = mass /molar mass of 1-hexene

=( 0.65 g / 84.16 g/mol)

=0.007723 mol.

#Amount of HBr = 5.0 mL

# Molarity of HBr = 1.8 M

We know that Molarity ,

Molarity = (number of moles /per liter of solution )

By rearranging,

Number of moles of HBr = Molarity x per liter of solution

= 1.8 M x (5.0 /1000)L

= 0.009 mol

So molar ratio of HBr against 1-hexene:

= moles of HBr /moles of 1-hexene

= (0.009mol /0.007723 mol)

= 1.16 equivalent, so moles of 1-hexene is smaller than moles of HBr , so 1-hexene treated as limiting reagent .

Product -

Molar mass of product = 165.07 g/mol

Theoretical mass of product = moles of 1-hexene x molar mass of product

=( 0.007723 mol x 165.07 g/mol)

=1.27 g of product (theoretical mass )

2) Reaction -2

3,3-dimethyl-1-butene + HBr  $\LARGE \rightarrow$ since all products have same mass , so yield calculated it.

#molar mass of 3,3-dimethyl-1-butene = 84.16 g/mol

#density =0.65 g/ml

#mass = volume x density

=1.0 mL x 0.65 g/mL

=0.65 g

# moles of 3,3-dimethyl-1-butene = (0.65 g/ 84.16 g/mol)

= 0.007723 mol

# moles of HBr = Molarity x volume of solution in liter

= 1.8 M x (5.0/1000) L

= 0.009 mol of HBr

Moles of HBr $\LARGE >$ Moles of 3,3-dimethyl-1-butene

So moles of 3,3-dimethyl-1-butene is limiting reagent

3) product mass (theoretical )

= moles of 3,3-dimethyl-1-butene x molar mass of 3.3-dimethyl-1-butene

=( 0.007723 mol x 165.07 g/mol)

= 1.27 g of product