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Nookie1986 [14]
3 years ago
7

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it

is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction2COF2(g)⇌CO2(g)+CF4(g), Kc=5.70If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?
Chemistry
1 answer:
Anna11 [10]3 years ago
7 0

Answer:

[COF₂] = 0.346M

Explanation:

For the reaction:

2COF₂(g) ⇌ CO₂(g) + CF₄(g)

Kc = 5.70 is defined as:

Kc = [CO₂] [CF₄] / [COF₂]² = 5.70 <em>(1)</em>

Equilibrium concentrations of each compound after addition of 2.00M COF₂ will be:

[COF₂] : 2.00M - 2x

[CO₂] : x

[CF₄] : x

Replacing in (1):

5.70 =  [X] [X] / [2-2x]²

22.8 - 45.6x + 22.8x² = x²

0 = -21.8x² + 45.6x - 22.8

Solving for x:

X = 1.265 <em>-False answer, will produce negative concentrations-</em>

<em>X = 0.827.</em>

Replaing, molar concentration of COF₂ is:

[COF₂] : 2.00M - 2×0.827 = <em>0.346M</em>

I hope it helps!

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3.75 moles CO2 (6.023 x 10^23 molecules /mole) = 2.26 x 10^24 molecules CO2

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Ammonia (NH3) reacts with oxygen to form nitric oxide (NO) and water vapor: 4NH3 + 502 4NO + 6H2O b) When 20.0 g NH3 and 50.0 g
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Answer: a) . Ammonia is the limiting reagent

b. Oxygen is left over and 0.1375 g of oxygen is left over.

c. The theoretical yield of NO is 35.29 g.

d. The theoretical yield of H_2O is 31.74 g.

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

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Given mass of ammonia = 20.0 g

Molar mass of ammonia = 17.031 g/mol

Putting values in equation 1, we get:

\text{Moles of ammonia}=\frac{20.0g}{17.031g/mol}=1.17mol

For O_2

Given mass of oxygen gas = 50.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{50.0g}{32g/mol}=1.6mol

The chemical equation for the reaction is

4NH_3+5O_2\rightarrow 4NO+6H_2O

By Stoichiometry of the reaction:

4 moles of ammonia reacts with = 5 moles of oxygen

So 1.17 moles of ammonia will react with = \frac{5}{4}\times 1.17=1.4625mol of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent and (1.6-1.4625)= 0.1375 g of oxygen is left unreacted.

Thus ammonia is considered as a limiting reagent because it limits the formation of product.

1. By Stoichiometry of the reaction:

4 moles of ammonia produces = 4 moles of NO

1.17 moles of ammonia will produce = \frac{4}{4}\times 1.17=1.17moles of NO

Mass of NO=moles\times {\text{Molar Mass}}=1.17\times 30=35.29g

Thus Theoretical yield of NO is 35.29 grams.

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4 moles of ammonia produces = 6 moles of H_2O

1.2 moles of ammonia will produce = \frac{6}{4}\times 1.2=1.8moles of H_2O

Mass of H_2O=moles\times {\text{Molar Mass}}=1.8\times 18.015=31.74g  H_2O

Thus Theoretical yield of H_2O is 31.74 grams.

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