Chemistry, 24.03.2020 20:15 soccerhannah290

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

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Answer from: freshysans4

$\large \boxed{1.6}$

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

$\text{Moles of R} = \text{3.0 g R} \times \dfrac{\text{1 mol R}}{\text{179.05 g R}} = \text{0.0168 mol R}$

(b) NaOH

(i) Mass of solution

$\text{Mass} = \text{25 mL} \times \dfrac{\text{1.11 g}}{\text{1 mL}} = \text{27.8 g}$

(ii) Mass of NaOH

$\text{Mass of NaOH} = \text{27.8 g solution} \times \dfrac{\text{10 g NaOH}}{\text{100 g solution}} = \text{2.78 g NaOH}$

(iii) Moles of NaOH

$\text{Moles of NaOH} = \text{2.78 g NaOH} \times \dfrac{\text{1 mol NaOH}}{\text{40 g NaOH}} = \text{0.0694 mol NaOH}$

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} \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 $\large \boxed{\textbf{1.6 g}}$}$

Answer from: Quest

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

Answer from: Quest

c. is the answer

by love gaming

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The theoretical yield of 1,2-epoxycyclohexane is grams, when starting with 3.0 grams of trans-2-bro...

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