WHO CAN SOLVE THIS MOLE question???

[The Ultimate]

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.

 

[XPFMember]

I will try,just give me sometime,please!!!

 

[The Ultimate]

oh please take your own time.......i can wait!!!
I tried solving it....but couldn't!!

 

[XPFMember]

btw do u know the ans. or not i mean the final ans.???coz question is tricky!!what if i do it in a wrong way,who knows if it is correct or not? :?

 

[The Ultimate]

sorry , no i do not know the answer...... :-(

 

[XPFMember]

if u dont mind plz tell where did u get the question from??I did it but i am sure my ans. is wrong cuz i am getting 0.44 which is for sure wrong,bcuz i know i am missing smthng but let me try to find out my mistake,anyway plz do tell......

 

[shalado95]

I would like to knw it as well! Eeee... It's creeping me out! :?

 

[XPFMember]

 

[The Ultimate]

idk my teacher gave it in a test ......... :-( and that was last year

 

[XPFMember]

hmmm...only this much info in the question...no steps or...anyway u too keep trying and we will do our best too!!

 

[shalado95]

Hey Math_angel... I can't seem to send u a private message! :?
But anyway... hows the prep going?

 

[XPFMember]

read this PM System down

 

[Nibz]

@MathAngel, how did you solve Chem's problem? You are the Angel of Maths, no.

 

[XPFMember]

cuz chemistry is one of my favourite subjects too..

 

[Nibz]

Why did 'nickname' yourself MathAngel then?

 

[XPFMember]

cuz maths is the most most most favourite and i can do that too Alhumdulilah,though i am not that perfect in chemistry i mean i can make mistakes in chemistry so cant compare it to maths,also i love maths,can u imagine exam (maths) is over long time back and i am still with mah questions!!

 

[XPFMember]

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

 

[The Ultimate]

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[XPFMember]

ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo.........................................ooooooops!!!!!!
Wa eyakum!!!!!!!!!!!!
btw hope u understood!!

 

[The Ultimate]

o.o sorry
nd yes i understood......................................................................