Question

The theoretical yield of 1,2-epoxycyclohexane is _______________ grams, when starting with 3.0 grams of trans-2-bromocyclohexanol. (Enter the number using 3 significant figures, i.e. 1.22) Given: 3.0 g of trans-2-bromocyclohexanol FW: 179.05 25 mL of 10% NaOH FW: 40 and density: 1.11 g/mL 1,2-epoxycyclohexane FW: 98.15

Answer 1

`\large \boxed{1.6}`

Step-by-step explanation:

1. Gather all the information in one place.

Mᵣ: 179.05 40 98.15

BrC₆H₁₀OH + NaOH ⟶ C₆H₁₀O + NaBr + H₂O

m/g: 3.0

V/mL: 25

C/%: 10

ρ/g·mL⁻¹ 1.11

Let's call BrC₆H₁₀OH "R" (reactant) and C₆H₁₀O "P" (product).

They have given us the amounts of two reactants and asked us to calculate the amount of product. This is a limiting reactant problem.

1. Calculate the moles of each reactant

(a) BrC₆H₁₀OH

`\text{Moles of R} = \text{3.0 g R} * \frac{\text{1 mol R}}{\text{179.05 g R}} = \text{0.0168 mol R}`

(b) NaOH

(i) Mass of solution

`\text{Mass} = \text{25 mL} * \frac{\text{1.11 g}}{\text{1 mL}} = \text{27.8 g}`

(ii) Mass of NaOH

`\text{Mass of NaOH} = \text{27.8 g solution} * \frac{\text{10 g NaOH}}{\text{100 g solution}} = \text{2.78 g NaOH}`

(iii) Moles of NaOH

`\text{Moles of NaOH} = \text{2.78 g NaOH} * \frac{\text{1 mol NaOH}}{\text{40 g NaOH}} = \text{0.0694 mol NaOH}`

3. Calculate the moles of C₆H₁₀O you can obtain from each reactant

The molar ratios are all 1:1, so

(a) 0.0168 mol R ⟶ 0.0168 mol P

(b) 0.0694 mol NaOH ⟶ 0.0694 mol P

R is the limiting reactant, because it gives fewer moles of P.

3. Calculate the theoretical yield.

`\text{Mass of P} = \text{0.0168 mol P} * \frac{\text{98.15 g P}}{\text{1 mol P}} = \textbf{1.6 g P}\\\\\text{The theoretical yield of 1,2-epoxycyclohexane is $\large \boxed{\textbf{1.6 g}}$}`