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Chemistry: Post your doubts here!

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aldehyde is formed under the condition of * distill over * not under heat under reflux. Just read the question carefully ...they have said that the acid formed is separated by distillation.
Did you get it or not ??? :unsure::unsure:

For discussion's sake, when the question says the product is collected by distillation. What do you think the distillate is?
 
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Which is my point of concern. Cause the acid should the higher point compared to alcohol.
The only sense i can make it it is they collect the alcohol distillate first, then collect the acid.

Exactly thats what i was thinking but did u notice the wording of question??? -_-

2.76 g of ethanol were mixed with an excess of aqueous acidified potassium dichromate(VI). The
reaction mixture was then boiled under reflux for one hour. The organic product was then
collected by distillation.
The yield of product was 75.0%.
What mass of product was collected?

There you see we have to consider the Acid formed not the Alcohol and who cares if we get acid or an Alcohol as a distillate.
 
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Exactly thats what i was thinking but did u notice the wording of question??? -_-

2.76 g of ethanol were mixed with an excess of aqueous acidified potassium dichromate(VI). The
reaction mixture was then boiled under reflux for one hour. The organic product was then
collected by distillation.
The yield of product was 75.0%.
What mass of product was collected?

There you see we have to consider the Acid formed not the Alcohol and who cares if we get acid or an Alcohol as a distillate.

Well, for MCQ, it doesn't really matter. Like i said, its just for discussion purpose, maybe will make a difference if its a theory paper where we need to design experiment procedure.
 
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Exactly thats what i was thinking but did u notice the wording of question??? -_-

2.76 g of ethanol were mixed with an excess of aqueous acidified potassium dichromate(VI). The
reaction mixture was then boiled under reflux for one hour. The organic product was then
collected by distillation.
The yield of product was 75.0%.
What mass of product was collected?

There you see we have to consider the Acid formed not the Alcohol and who cares if we get acid or an Alcohol as a distillate.
Mass differs because if aldehyde u get ans A if carboxylic acid ans is c
 
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Now make sense
Maybe a final point...

The reason why one would want to reflux alcohol with dichromate in the first place is that they wish to obtain acid instead of aldehyde.

So whatever "yield of product" mentioned later, the product should be understood to be acid.

Even if some aldehyde is indeed formed, its not included in the mass of product as it is not desired in the first place.

That might also be a reason why the yield is less than 100%, cause some became aldehyde instead of acid.
 
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Help with this please, ans is C
0451e-f6a6a0be-97a2-465a-98c9-9db2c87c924b.png
 
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Maybe a final point...

The reason why one would want to reflux alcohol with dichromate in the first place is that they wish to obtain acid instead of aldehyde.

So whatever "yield of product" mentioned later, the product should be understood to be acid.

Even if some aldehyde is indeed formed, its not included in the mass of product as it is not desired in the first place.

That might also be a reason why the yield is less than 100%, cause some became aldehyde instead of acid.
Thanks alot :)
 
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Exactly thats what i was thinking but did u notice the wording of question??? -_-

2.76 g of ethanol were mixed with an excess of aqueous acidified potassium dichromate(VI). The
reaction mixture was then boiled under reflux for one hour. The organic product was then
collected by distillation.
The yield of product was 75.0%.
What mass of product was collected?

There you see we have to consider the Acid formed not the Alcohol and who cares if we get acid or an Alcohol as a distillate.
Thanks for your help :)
 
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3 : 1 mole ratio of O₂ , so only 30 cm³ of oxygen react wid 10 cm³ of CS₂. Oxygen left is 20 cm³, CO₂ and SO₂ formed has volume 10cm³ n 20 cm³ respectively from mole ratio. Total volume of gas at end of reaction = 20 cm³ of unreacted O₂ + 10 cm³ CO₂ and 20 cm³ SO₂ :)
CO₂ n SO₂ r both acidic, so they both react completely wid NaOH, n all dat is left is da 20 cm³ of unreacted O₂ :)
 
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