Chemistry: Post your doubts here!

[Mimick]

somebody please explain why do we have to multiply moles of H2 by 2? I have my mock tomorrow

Send me a proper pic.. I need the data

 

[princess Anu]

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[My Name]

How come its C ?? ( I choose A)

 

[The Chill Master]

View attachment 51450
How come its C ?? ( I choose A)

In A it's propane and the reactions will continue till conditions are maintained so you may get 1-chloropropane or 2-chloropropane.(Keep this in mind if we have no other better options we will go with this)
In B,C & D you can eliminate B & D since it is dilute you have less chances of getting more yield you are left with C & A.
With C & A you will get same amount of yield according to me but you have a greater chance of obtaining a same product with C.so the ans is C

 

[princess Anu]

can somebody do Q 4

 

[awesomaholic101]

yes. I didn't get why u multiplied the moles of H2 by 2?

Initially, they are adding 1 mol each of H2O and CO, yea? And the initial no. of moles of products is 0. That is a total of 2 moles.
You must know that, whatever the position of equilibrium, the total no. of moles in this case is 2 as the ratio of products to reactants is 1:1, according to the given balanced equation.

Now, given that, at equilibrium, 33.3% of the total moles (2) is H2 (we can take volume as moles as all are gases) ...
n(H2) = 33/100 * 2 = 0.666

Then comes the rest of the calculations.

 

[My Name]

http://onlineexamhelp.com/wp-content/uploads/2013/10/9701_s13_qp_22.pdf

Question 5 part d)i) how is it ethanoic acid?

http://onlineexamhelp.com/wp-content/uploads/2013/10/9701_s12_qp_22.pdf

Question3

HO2CCH2CH(OH)CO2H <--That’s the acid given

and then in part B the reaction is heat with CH3CO2H/H+ <--This to is an acid...

so the Mark scheme says HO2CCH2CH(OH)CO2H (with CH3CO2H/H+)-->

How did that form?

 

[Metanoia]

Initially, they are adding 1 mol each of H2O and CO, yea? And the initial no. of moles of products is 0. That is a total of 2 moles.
You must know that, whatever the position of equilibrium, the total no. of moles will remain constant; in this case 2.

Now, given that, at equilibrium, 33.3% of the total moles (2) is H2 (we can take volume as moles as all are gases) ...
n(H2) = 33/100 * 2 = 0.666

Then comes the rest of the calculations.

Do be very careful of the assumption that the total moles would remain constant, conceptually, it is wrong.
It works in this unique question as the sum of the stoichiometric coefficients are the same on both sides.

Using an example, hydrogen reacts with oxygen to form water

2H2 + O2 --> 2H2O
Taking the reaction to completion, using 2 mole of H2 and 1 moles of H2 will not give us (1+2) moles of H2O.

 

[Metanoia]

can somebody solve thissView attachment 51422

 

[awesomaholic101]

Do be very careful of the assumption that the total moles would remain constant, conceptually, it is wrong.
It works in this unique question as the sum of the stoichiometric coefficients are the same on both sides.

Using an example, pretending that Nitrogen and Hydrogen reacts irreversibly to form ammonia

N2 + 3H2 --> 2NH3

Using 1 mole of N2 and 3 moles of H2 will not give us (1+3) moles of NH3.

Yes, it depends on the ratio of products: reactants which is 1:1 in that particular example ... we must calculate according to the balanced equation.

 

[awesomaholic101]

http://onlineexamhelp.com/wp-content/uploads/2013/10/9701_s13_qp_22.pdf

Question 5 part d)i) how is it ethanoic acid?

http://onlineexamhelp.com/wp-content/uploads/2013/10/9701_s12_qp_22.pdf

Question3

HO2CCH2CH(OH)CO2H <--That’s the acid given

and then in part B the reaction is heat with CH3CO2H/H+ <--This to is an acid...

so the Mark scheme says HO2CCH2CH(OH)CO2H (with CH3CO2H/H+)-->View attachment 51470

How did that form?

I remember doing that question with R, but i dont remember how ... in shaa Allah i will try to get back to u on that.

For ur second question ...

HO2CCH2CH(OH)CO2H <--- This is a dioic acid with a hydroxyl group present
CH3CO2H/H+ <---- This is an acid

So naturally, acid groups of both reactants cannot react. Therefore the hydroxyl group of HO2CCH2CH(OH)CO2H (cut H from OH grp) and acid grp of CH3CO2H react (just cut off OH) to give you ... (after cutting, join the O from 1st molecule and C frm 2nd molecule) HO2CCH2CH(OCOCH3)CO2H <--- thats the given molecule in ms

my explanation is dumb i kno ... but u will understand if u follow the steps by drawing the displayed formulae of each molecule

 

[Wkhan860]

http://onlineexamhelp.com/wp-content/uploads/2013/10/9701_s13_qp_22.pdf

Question 5 part d)i) how is it ethanoic acid?

http://onlineexamhelp.com/wp-content/uploads/2013/10/9701_s12_qp_22.pdf

Question3

HO2CCH2CH(OH)CO2H <--That’s the acid given

and then in part B the reaction is heat with CH3CO2H/H+ <--This to is an acid...

so the Mark scheme says HO2CCH2CH(OH)CO2H (with CH3CO2H/H+)-->View attachment 51470

How did that form?

NAHCO thngy teacts only wth ethanoic acid (its an acid base reacn) to give CO2
It neither reacts wth a carbonyl comp nor wth ethanol so its an acid

 

[The Chill Master]

http://onlineexamhelp.com/wp-content/uploads/2013/10/9701_s13_qp_22.pdf

Question 5 part d)i) how is it ethanoic acid?

http://onlineexamhelp.com/wp-content/uploads/2013/10/9701_s12_qp_22.pdf

Question3

HO2CCH2CH(OH)CO2H <--That’s the acid given

and then in part B the reaction is heat with CH3CO2H/H+ <--This to is an acid...

so the Mark scheme says HO2CCH2CH(OH)CO2H (with CH3CO2H/H+)-->View attachment 51470

How did that form?

For the first one you need to have a glance at your structures and think what will be left when each ester is hydrolyzed.For eg when i drew the isomers I got my answer as W.The formula for W was CH3CO2CH2CH3 now in this case you are getting CH3CO2H as your acid and instead of an Alchol I am getting an Alkane CH3CH3. There you go it's not an aldehyde nor a ketone.

 

[My Name]

NAHCO thngy teacts only wth ethanoic acid (its an acid base reacn) to give CO2
It neither reacts wth a carbonyl comp nor wth ethanol so its an acid

I know its an ACID but why ethanoic was my question.
Thnx anyway

For the first one you need to have a glance at your structures and think what will be left when each ester is hydrolyzed.For eg when i drew the isomers I got my answer as W.The formula for W was CH3CO2CH2CH3 now in this case you are getting CH3CO2H as your acid and instead of an Alchol I am getting an Alkane CH3CH3. There you go it's not an aldehyde nor a ketone.

I didnt get you

 

[Pink Flower]

Which is more acidic, 2-chloropropanoic acid or 3-chloropropanoic acid? Explain your answer.
I think 2-chloropropanoic acid because it will have more electron withdrawing effect. Am I correct?

 

[Metanoia]

Which is more acidic, 2-chloropropanoic acid or 3-chloropropanoic acid? Explain your answer.
I think 2-chloropropanoic acid because it will have more electron withdrawing effect. Am I correct?

Yes. 2-chloropropanoic acid is stronger, this is due to the proximity of the electron withdrawing Cl to the COOH group.

You can explain it either in terms of the weakened O-H bond before losing the proton or the stability of the anion after losing the proton.

weakened O-H bond
The O-H bond is more polarized, thus the O-H bond is more easily broken to release H+.

stability of the anion
After losing a proton, a COO- charge is formed. The negative charge is more effectively dispersed by the chlorine atom that is nearer to the COO-.

 

[Metanoia]

I know its an ACID but why ethanoic was my question.
Thnx

Did you find out what is the formula of Q?

 

[My Name]

Did you find out what is the formula of Q?

molecular formula= C4H8O2

 

[Metanoia]

molecular formula= C4H8O2

That is the ester right?

Then the question states that Q is hydrolyzed with sulfuric acid

C4H8O2 --> acid + alcohol

Then the mixture was oxidized to obtained a single product, R, in other words, the alcohol in the above reaction is oxidized to become the same compound as the acid.

It simply means the alcohol has the same number of carbons as the acid.

We have 4 carbons to divide equally among the two....so the alcohol was ethanol and the acid was ethanoic acid.

When this mixture was oxidized, all we end up with is a sample of ethanoic acid.

 

[princess Anu]

View attachment 51471

What if THIS equation had 2 moles on one side and 3 on other..?
We couldnot solve it right?