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sveticcg [70]
3 years ago
6

The equilibrium constant has been estimated to be 0.12 at 25 °C. If you had originally placed 0.069 mol of cyclohexane in a 2.8

L flask, what would be the concentrations of cyclohexane and methylcyclopentane when equilibrium is established?
Chemistry
1 answer:
scZoUnD [109]3 years ago
4 0

Answer: Concentrations of cyclohexane and methylcyclopentane at equilibrium are 0.0223 M and 0.0027 M respectively

Explanation:

Moles of cyclohexane = 0.069 mole

Volume of solution = 2.8 L

Initial concentration of cyclohexane =\frac{moles}{Volume}=\frac{0.069}{2.8}=0.025M

The given balanced equilibrium reaction is,

                            cyclohexane  ⇔  methylcyclopentane

Initial conc.                 0.025 M           0

At eqm. conc.       (0.025-x)M       (x) M

The expression for equilibrium constant for this reaction will be,

K= methylcyclopentane / cyclohexane

Now put all the given values in this expression, we get :

0.12=\frac{(x)}{(0.025-x)}

By solving the term 'x', we get :

x =  0.0027

Concentration of cyclohexane at equilibrium = (0.025-x ) M = (0.025-0.0027) M = 0.0223 M

Concentration of methylcyclopentane at equilibrium = (x ) M = (0.0027) M

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2 years ago
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2 years ago
When methane, CH4, is combusted, it produces carbon dioxide, CO₂, according to the unbalanced equation: CH4 +0₂ - CO₂ + H₂O.
nevsk [136]

The balanced equation will be CH_4 + 2O_2 --- > CO_2 + 2H_2O

<h3>Stoichiometric problems</h3>

The balanced equation of the reaction would be as follows:

CH_4 + 2O_2 --- > CO_2 + 2H_2O

The mole ratio of methane to carbon dioxide is 1:1.

10 grams of methane will give 10/16 = 0.625 moles

0.625 moles of carbon dioxide would give 0.625 x 44.01 = 27.506 grams.

Thus, 10 grams of methane will produce approximately 27 grams of carbon dioxide stoichiometrically.

The reaction obeys the law of conservation of mass because the atoms of all the elements before and after the reaction are balanced.

More on stoichiometric problems can be found here: brainly.com/question/14465605

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3 0
2 years ago
If 18.1 g of ammonia is added to 27.2 g of oxygen gas, how many grams of excess reactant is remaining once the reaction has gone
GREYUIT [131]

Answer:

m of NH3 = 6.46 g

Explanation:

First, in order to know the limiting and excess reactant, we need to write and balance the equation that is taking place:

NH₃ + O₂ ---------> NO + H₂O

Now, let's balance the equation:

4NH₃ + 5O₂ ---------> 4NO + 6H₂O

Now that we have the balanced equation, let's see which reactant is in excess. To know that, let's calculate the moles of each reactant using the molar mass:

MM NH3 = 17 g/mol

MM O2 = 32 g/mol

moles NH3 = 18.1 / 17 = 1.06 moles

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Now, let's compare these moles with the theorical moles that the balanced equation gave:

4 moles NH3 --------> 5 moles O2

1.06 moles ----------> X

X = 1.06 * 5 / 4 = 1.325 moles of O2

These means in order to  NH3 completely reacts with O2, it needs 1.325 moles of O2, which we don't have it. We only have 0.85 moles of O2, therefore, the limiting reactant is the O2 and the excess is NH3.

Now, let's see how many grams in excess we have left after the reaction is complete.

4 moles NH3 --------> 5 moles O2

X moles NH3 ----------> 0.85 moles

X = 0.85 * 4 / 5 = 0.68 moles of NH3

This means that 0.85 moles of O2 will react with only 0.68 moles of NH3, and we have 1.06 so, the remaining moles are:

moles remaining of NH3 = 1.06 - 0.68 = 0.38 moles

Finally the mass:

m = 0.38 * 17

<em>m = 6.46 g of NH3</em>

8 0
3 years ago
Which of the following best describes carbon dioxide?
sdas [7]

Answer:   B. Compound

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Although the phase of the carbon dioxide is gaseous that is single phase, stll it is not considered as the Homogeneous mixture because in carbon dioxide , both the elements carbon and oxygen reacts with each other chemically to form the gas. This doesnot happen in te mixture.

Thus carbon dioxide is considered as the compound.

3 0
3 years ago
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