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Homework Help: Stoichiometry - 0.7g of Na2CO3.xH2O were dissolved in water

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data
    0.7g of Na2CO3.xH2O were dissolved in water and the volume was made to 100 ml, 20 ml of this solution required 19.8 ml of N/10 HCl for complete neutralization. The value of x ?

    2. Relevant equations

    Na2CO3 + 2HCl = 2NaCl + H2O + CO2

    3. The attempt at a solution

    I have taken the Normality of sodium carbonate solution as 0.07 N. Since the solution is dissolved in water and not in any other solution, the number of molecules of water in sodium carbonate should not matter for calculating the normality of the solution. This requires 19.8 ml of 0.1 N HCl. I have tried balancing the equivalents, i.e the equivalent of sodium carbonate is Normality x volume = 0.07 * 20 = 0.14. The equivalents of HCl is 0.1 * 19.8 = 1.98.

    Where does the "x" come in ?
    Last edited by a moderator: Jul 9, 2011
  2. jcsd
  3. Jul 9, 2011 #2
    Re: Stoichiometry

    "I have taken the Normality of sodium carbonate solution as 0.07 N." That's your mistake. The problem statement doesn't tell you that. It says 0.7 GRAMS of Na2CO3.xH2O were dissolved, not 0.7 MOLES. In order to solve this problem, you need to figure out how many moles of Na2CO3.xH2O 0.7g is.
  4. Jul 9, 2011 #3
    Re: Stoichiometry

    OK, so the number of moles of Na2CO3.xH2O is 0.7 / 106 + 18x. What next?
  5. Jul 9, 2011 #4


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    Re: Stoichiometry

    The number of moles of Na2CO3.xH2O is 0.7/(106 + 18x).

    Not a pedantic difference.

    OK, your eqation

    Na2CO3 + 2HCl = 2NaCl + H2O + CO2

    is telling you one mole of Na2CO3 reacts with how many moles of HCl?

    From the problem statement how many moles of HCl have reacted? Therefore how many moles did you have there?
  6. Jul 9, 2011 #5
    Re: Stoichiometry

    one mole of HCl is 36.5 g, which means one-half mole of HCl has reacted with the aq. sodium carbonate. Since normally one mole of sodium carbonate reacts with two moles of HCl, in this case it means one mole of Na2CO3 has reacted with half a mole of HCl.

    This means that Na2CO3 is one fourth of a mole? i.e 26.5 g ? But what we have here is only 0.7 g that too Na2CO3.xH2O. How to relate that to the figure 0.7 / 106 + 18x ?
  7. Jul 9, 2011 #6
    Re: Stoichiometry

    Wrong again, Buffalo Jack! 19.8 is not the amount of HCl that reacted with the carbonate. It's the amount of N/10 HCl solution that reacted with carbonate.

    You've got to pay attention to the details. As epenguin says, "Not a pedantic difference."
  8. Jul 9, 2011 #7
    Re: Stoichiometry

    I am complete newbie to stoichiometry, so please bear with me. I have been struggling with this problem for a long time but I finally figured out this much, though I don't know if this is correct:

    Since 0.7 grams are dissolved in 100 ml of water, in 20 ml the weight would be 0.7/5 = 0.14 g. This is neutralized by 19.8 ml of 0.1 N HCl i.e. 1.98 ml is required for N HCl.

    no. of moles of hydrated sodium carbonate = given mass / mol mass = 0.14 /106 + 18x

    How do I proceed from here?
  9. Jul 9, 2011 #8
    Re: Stoichiometry

    How many moles of bicarbonate would be neutralized by 1.98 ml of N HCl?
  10. Jul 9, 2011 #9
    Re: Stoichiometry

    I should probably equate 1.98/1000 and 0.14 / 106+18x, though I am not sure. If I solve this, I get approximately 2, which is the right answer. Is my approach right?
  11. Jul 9, 2011 #10
    Re: Stoichiometry

    Umm, no. If you equate 1.98/1000 and 0.14 / 106+18x, you get x = -2. (That's a minus sign.) Unless your bicarbonate is complexed with anti-matter, that is probably not the answer you're looking for.

    Take it one step at a time. How many moles of bicarbonate would be neutralized by 1.98 ml of N HCl?
  12. Jul 9, 2011 #11
    Re: Stoichiometry

    I give up. Please tell me how to go about? I have spent too long a time on this one problem.

    I know 106 g i.e. one mole of Na2CO3 is neutralized by 73 g, i.e 2 moles of HCl. I don't know how to find out how many moles react with 1.98 ml of HCl. This is my first week at Chem.
  13. Jul 9, 2011 #12
    Re: Stoichiometry

    In that case, I also give up. Maybe someone else will tell you the answer.
  14. Jul 9, 2011 #13
    Re: Stoichiometry

    OK, so be it.

    Thanks for your patience anyway.
  15. Jul 10, 2011 #14


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    OK since you are confused but have done some work I will give you my way to answer (the calc. could be set out in different ways according to preference).

    But, I don’t know why people get into these confusions (as opposed to making errors which I do not exclude I have done myself) however the thing to bear in mind is it is all basically quite simple. The point is that molecules react in simple ratios, most often 1:1 or 1:2, sometimes you will see a 3, 4, or 5. But the weights or masses, the grams, of the things, atoms or molecules reacting are different and not simple. So that has to be taken account of in the calcs. E.g. SO2 contains 1 atom of sulphur two of oxygen. Simple. But O weighs 16 to S’s 35.5 Result SO2 contains 35.5g sulphur and 32g oxygen, not an apparently simple ratio. But convert it back to moles using atomic masses and it is simple. Use coloured blobs to get used to the idea, but that is what your chemical equations like your

    Na2CO3 + 2HCl = 2NaCl + H2O + CO2

    are saying.

    Then there are complications of volumes, samples, etc. again usually just arithmetic problems of proportions; these are not just invented to complicate but usually are part of practical situations; your problem is typical

    In the experiment you find the carbonate in the sample reacts with 19.8 ml 0.1 M HCl.

    That is 19.8 X 0.1/1000 = 0.00198 moles HCl

    According to your equation 2 moles of HCl reacts with 1 mole of carbonate.

    Therefore there were 0.00099 moles of carbonate in the sample reacted. (We could almost afford to round off but it might tell us something extra if we don’t.)

    These moles were contained, taking into account the sampling, in 0.7/5 g of carbonate, i.e. 0.14 g carbonate.

    So 0.14 g of this stuff is 0.00099 moles.

    So 1 mole of Na2CO3.xH2O is 0.14/0.00099 g =141.1 g.

    In other words the molecular mass of Na2HCO3.xH2O is 141.1, by experimental measurement.

    The molecular mass of Na2HCO3 with no water of crystallisation is, from atomic weights, again I use the accurate ones just in case useful, 22.99X2 + 12.01 + 16.00X3 = 105.99

    So of a mole of Na2CO3.xH2O 105.99 g is Na2CO3 and the rest, 141.1 – 105.99 = 35.11 g is H2O. Divide by the molecular mass of H2O, 18.016, I get 1.95, close to 2.

    x = 2 would be a good enough answer to your problem.

    Perhaps someone will check my calcs. but to get near a whole number is already reassuring it is probably right. Thing is I have never heard of such a crystal form for Na2CO3.xH2O before. I have heard of x = 1 and x = 10. An error somewhere??

    For many stoichiometric questions the nearly whole numbers to which answers approximate are good and useful enough, not only for students. It is an advantage of stoichiometry that approximate numbers are still useful.

    But there is also an advantage in having done the calc more accurately. The discrepancy is 2.5%, that would correspond to an error of 0.25 ml in the titration. So the question to ask would be is that outside experimental error? I’d say a bunch of students between them could do better than that, but it could be checked if you were there. If it is outside experimental error you have to think e.g. maybe the crystals have lost water – sodium carbonate crystals do do that - or something like that.

    If you look your book I would be surprised if there are not how to do some similar calculations explained. E.g calculations of purity are rather similar.
    Last edited: Jul 10, 2011
  16. Jul 11, 2011 #15
    epenguin, thank you so much for your detailed explanation and working. When I last came here, trying to work out this problem, I was tired and confused and no wonder I couldn't make out what was being told.

    As I see it, it was actually fairly simple. I later went back to the problem and solved it in a slightly different way.

    I got the 141.1 and equated it with 106 + 18x and got x as approximately 2.

    Thanks again.:smile:
  17. Jul 12, 2011 #16


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    :approve: Good. It must often happen that people clear up their problems by themselves before they see the help, actually the very writing out of the problem often helps towards solving. But it is good to come back and say so, and even state the solution as you have (because in some case they may still be wrong). Otherwise the helper is left high and dry not knowing whether his post was or could have been useful.
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