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3 years ago

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While performing the Acid-catalyzed Hydrolysis of Epoxides experiment you used a solution containing 1.84 mL of cyclohexene oxid

e. After the reaction was completed and worked up, you obtained 0.739 g of 1,2-cyclohexane diol. Calculate the moles of starting material used.

1 answer:

Elenna [48]3 years ago

3 0

Answer:

The moles of starting material is $1.8212 \times 10^{-2}$ mole

Explanation:

Given:

Cyclohexene oxide material is used

Density = 0.97 $\frac{g}{mL}$

Volume = 1.84 mL

Formula of Cyclohexene oxide = $C_{6} H_{10} O$

Molar mass is given by,

$C = 6 \times 12 = 72$

$H = 10 \times 1 = 10$

$O = 1 \times 16 = 16$

⇒ $m' = 96\frac{g}{mol}$

Mass $m = 0.97 \times 1.84 = 1.7848$ $g$

Moles is given by,

$n = \frac{m}{m'}$

$n = \frac{1.7848}{98}$

$n = 1.8212 \times 10^{-2}$ mol

Therefore, the moles of starting material is $1.8212 \times 10^{-2}$ mole

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The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

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Compound Amount

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<u>Answer:</u> The value of equilibrium constant for given equation is $1.0\times 10^{-4}$

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

<u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M$

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Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M$

For the given chemical equation:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

The expression of equilibrium constant for above equation follows:

$K_{eq}=\frac{[H_2O]^3}{[H_2]^3}$

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}$

Hence, the value of equilibrium constant for given equation is $1.0\times 10^{-4}$

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To calculate activation energy of the reaction, we use Arrhenius equation for two different temperatures, which is:

$\ln(\frac{K_{317K}}{K_{278K}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{317K}$ = equilibrium constant at 317 K = $3.050\times 10^{8}M^{-1}s^{-1}$

$K_{278K}$ = equilibrium constant at 278 K = $3.600\times 10^{7}M^{-1}s^{-1}$

$E_a$ = Activation energy = ?

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = 278 K

$T_2$ = final temperature = 317 K

Putting values in above equation, we get:

$\ln(\frac{3.050\times 10^8}{3.600\times 10^{7}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{278}-\frac{1}{317}]\\\\E_a=40143.3J/mol=40.143kJ/mol$

Hence, the activation energy for the reaction is 40.143 kJ/mol

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