Answer:
[tex]\large \boxed{1.6}[/tex]
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
[tex]\text{Moles of R} = \text{3.0 g R} \times \dfrac{\text{1 mol R}}{\text{179.05 g R}} = \text{0.0168 mol R}[/tex]
(b) NaOH
(i) Mass of solution
[tex]\text{Mass} = \text{25 mL} \times \dfrac{\text{1.11 g}}{\text{1 mL}} = \text{27.8 g}[/tex]
(ii) Mass of NaOH
[tex]\text{Mass of NaOH} = \text{27.8 g solution} \times \dfrac{\text{10 g NaOH}}{\text{100 g solution}} = \text{2.78 g NaOH}[/tex]
(iii) Moles of NaOH
[tex]\text{Moles of NaOH} = \text{2.78 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40 g NaOH}} = \text{0.0694 mol NaOH}[/tex]
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.
[tex]\text{Mass of P} = \text{0.0168 mol P} \times \dfrac{\text{98.15 g P}}{\text{1 mol P}} = \textbf{1.6 g P}\\\\\text{The theoretical yield of 1,2-epoxycyclohexane is $\large \boxed{\textbf{1.6 g}}$}[/tex]
