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madreJ [45]
3 years ago
10

Answer the following question: Ethanol, C2H5OH, is considered clean fuel because it burns in oxygen to produce carbon dioxide an

d water with few trace pollutants. If 150.0 g of H2O are produced during the combustion of ethanol, how many grams of ethanol were present at the beginning of the reaction?
Chemistry
1 answer:
Delvig [45]3 years ago
7 0

Answer:

127.88 grams of ethanol were present at the beginning of the reaction

Explanation:

Firstly, let's make the combustion reaction:

C₂H₅OH  +  3O₂  →  2CO₂  +  3H₂O

By 1 mol of ethanol, you can make 3 mole of water.

Mole of water = Water mass / Molar mass

150g / 18g/m = 8.3 mole

3 mole of water came from 1 mol of ethanol

8.3 mole came from (8.3 .1)/3 = 2.78 mole of ethanol

Molar mass ethanol = 46 g/m

Mole . molar mass = mass

2.78 m .  46g/m = 127.88 g

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While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e
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The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

<u>Answer:</u> The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                  1.8                4.7

<u>At eqllm:</u>           1.8-x             4.7-x

Evaluating the value of 'x'

\Rightarrow (1.8-x)=1.16\\\\x=0.64

The expression of K_p for above equation follows:

K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}

p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm

Putting values in above expression, we get:

K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

                     CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)

<u>Initial:</u>                2.36             4.06               0.64

<u>At eqllm:</u>           2.36-x        4.06-x             0.64+x

Putting value in the equilibrium constant expression, we get:

0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

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