WHO CAN SOLVE THIS MOLE question???

[SuperXDE]

Does anybody know how to answer this?

Sodium carbonate crystals have the formula Na2CO3 n H20
A 27.823g sample of crystals was dissolved in water. 25.0 cm3 of this solution was reacted with 48.8 cm3 of 0.1 mol/dm3 hydrochloric acid.
Find n in the formula of crystals.

There is a missing element yet in the Question , How much water was the 27.823g sample dissolved in? Well , supposing that it is 1000cm3 , It is the answer Math_Angel worked out. however the volume of the solute is required . But I tried to solve it with my hand ... it appears that something is wrong with the question mentioned above.

Here is the correct question , found it on WikiAnswers

27.82 grams of hydrated sodium carbonate crystals was dissolved in water and made up to 1 liter 0.0025 of this solution was neutralized by 0.0488 of hydrochloric acid of concentration 0.01 molar?

 

[Saturation]

Tough qs! btw has this ever come in a past papers? and is it likely that they could ask something like this!? :S

 

[shalado95]

Tough qs! btw has this ever come in a past papers? and is it likely that they could ask something like this!? :S

Complicated mole calculation questions have come time to time... but I haven't come across any thing like the above question. :? Though u nvr knw... so u need to have a back up plan any way!

Does anybody know how to answer this?

Sodium carbonate crystals have the formula Na2CO3 n H20
A 27.823g sample of crystals was dissolved in water. 25.0 cm3 of this solution was reacted with 48.8 cm3 of 0.1 mol/dm3 hydrochloric acid.
Find n in the formula of crystals.

There is a missing element yet in the Question , How much water was the 27.823g sample dissolved in? Well , supposing that it is 1000cm3 , It is the answer Math_Angel worked out. however the volume of the solute is required . But I tried to solve it with my hand ... it appears that something is wrong with the question mentioned above.

Here is the correct question , found it on WikiAnswers

27.82 grams of hydrated sodium carbonate crystals was dissolved in water and made up to 1 liter 0.0025 of this solution was neutralized by 0.0488 of hydrochloric acid of concentration 0.01 molar?

Even though the whole question is mentioned n the workings done... I still can't work it out!! :?


though JazakAllah Khairan!

 

[shalado95]

eee... I just saw the answer on the site.
It was quiet complicated!! :?

 

[niassu]

DONE!!!(btw i didnt do it took help from someone )
First work out how many moles of HCl were used.
One carbonate reacts with one acid ... so this is the moles of carbonate present in 25cm3.
Scale this up to the moles in 1dm3 and this is the number of moles in 27.82g.

The eqation is
Na2CO3 + 2HCl --> NaCl + CO2 + H2O
So TWO moles of acid react with one mole of Na2CO3



48.8 cm3 of 0.1M HCl = 0.00488moles

Na2CO3 + 2HCl --> NaCl + CO2 + H2O

therefore moles of Na2CO3 = 0.00488/2 = 0.00244moles

This is in 25cm3 therefore the moles in 1000cm3 = 0.00244/0.025 =0.0976moles

If the formula = Na2CO3.nH2O

Then the neutralization/reaction has measured only the Na2CO3

Therefore the mass of Na2CO3 = RMM x no of moles = 106 x 0.0976 = 10.3456g

The remaining mass must be due to water = 27.823 - 10.3456 = 17.4774g

RMM of water = 18 therefore this is equivalent to 17.4774/18 moles = 0.971

Thus the mole ratio of Na2CO3 to water in the original compound = 0.096 : 0.971

or approximately 1 :10

The formula is therefore Na2CO3.10H2O

are u sure that this is the water of crysatls and no water of neautralisation is in it ? what about the water that dissolved the crystals>? Hope u cler my doubts

 

[SuperXDE]

Tough qs! btw has this ever come in a past papers? and is it likely that they could ask something like this!? :S

Complicated mole calculation questions have come time to time... but I haven't come across any thing like the above question. :? Though u nvr knw... so u need to have a back up plan any way!

Does anybody know how to answer this?

Sodium carbonate crystals have the formula Na2CO3 n H20
A 27.823g sample of crystals was dissolved in water. 25.0 cm3 of this solution was reacted with 48.8 cm3 of 0.1 mol/dm3 hydrochloric acid.
Find n in the formula of crystals.

There is a missing element yet in the Question , How much water was the 27.823g sample dissolved in? Well , supposing that it is 1000cm3 , It is the answer Math_Angel worked out. however the volume of the solute is required . But I tried to solve it with my hand ... it appears that something is wrong with the question mentioned above.

Here is the correct question , found it on WikiAnswers

27.82 grams of hydrated sodium carbonate crystals was dissolved in water and made up to 1 liter 0.0025 of this solution was neutralized by 0.0488 of hydrochloric acid of concentration 0.01 molar?

Even though the whole question is mentioned n the workings done... I still can't work it out!! :?


though JazakAllah Khairan!

Wa Eyakoum ,

The Question is based on three points , Mole = Concentration × Volume , Mole = Mass / Relative Mass (Either Molecular or Atomic ) and Ratio between Reactants.

I didn't see you online for a while.

are u sure that this is the water of crysatls and no water of neautralisation is in it ? what about the water that dissolved the crystals>? Hope u cler my doubts

It is Water of Crystallisation , Water of Neutralisation is the product of Neutralisation reaction , on the other side of the hill.

 

[Aseel]

if there is equation plz write it

 

[XPFMember]

which equation do u need

 

[Aseel]

niassu your equation isnot balanced especially Na

 

[niassu]

but i didnot write any equations

 

[XPFMember]

ok i got what u need,it will be like this=>> Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O

 

[Aseel]

sorry niassu math angel wrote the equation and sodium isnot balanced.If a question like this come the equations will be written.My advise is to focus on the concept and leave you with that question.

 

[Aseel]

you have understood me math angel

 

[XPFMember]

 

[M.S.Princess93]

i still dont understand it

 

[SuperXDE]

i still dont understand it

It is all about Maths ... and the most important thing is mole questions is to focus on the Volume stuff if Gases , and the ratios between Masses , Volumes and Moles. ( One mole is equal to any other mole in volume , at r.t.p and normal pressure , 24litres. )