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mash [69]
3 years ago
12

A solution is made by mixing equal masses of methanol, CH 4 O , and ethanol, C 2 H 6 O . Determine the mole fraction of each com

ponent to at least three significant figures.
Chemistry
1 answer:
Fynjy0 [20]3 years ago
6 0

Answer:

Mole fraction of CH_4O = 0.58

Mole fraction of C_2H_6O = 0.42

Explanation:

Let the mass of CH_4O and C_2H_6O = x g

Molar mass of CH_4O = 33.035 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles_{CH_4O}= \frac{x\ g}{33.035\ g/mol}

Moles_{CH_4O}=\frac{x}{33.035}\ mol

Molar mass of C_2H_6O = 46.07 g/mol

Thus,

Moles= \frac{x\ g}{46.07\ g/mol}

Moles_{C_2H_6O}=\frac{x}{46.07}\ mol

So, according to definition of mole fraction:

Mole\ fraction\ of\ CH_4O=\frac {n_{CH_4O}}{n_{CH_4O}+n_{C_2H_6O}}

Mole\ fraction\ of\ CH_4O=\frac{\frac{x}{33.035}}{\frac{x}{33.035}+\frac{x}{46.07}}=0.58

Mole fraction of C_2H_6O = 1 - 0.58 = 0.42

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Answer:

A) N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g).

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C) 0.9 mol.

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Explanation:

Hello,

In this case, given the information, we proceed as follows:

A)

N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

B) For the calculation of Kc, we rate the equilibrium expression:

Kc=\frac{[NH_3]^2}{[N_2][H_2]^3}

Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent x, we have:

[NH_3]=0.6M=2*x

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Next, the concentrations of nitrogen and hydrogen at equilibrium are:

[N_2]=\frac{1.5mol}{1.5L}-x=1M-0.3M=0.7M

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C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.

D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.

Best regards.

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