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.
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.
