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Can you help? Chemistry F321 Water of Crystallisation watch

1. Stuck on the following question of the previous exam June 14 F321. We did it for our mock and I had absolutely had no idea, after being given a hard time about it in lesson I'm still stuck and was just wondering if anyone could help before I go back and ask?

Hydrated strontium chloride, SrCl2 . 6H2O, has a molar mass of 266.6 gmol^-1.
A student heats 5.332g of SrCl2 . 6H2O.
The SrCl2 . 6H2O loses SOME of its water of crystalisation forming 3.892g of a solid product.
Use the information above to dertermin the formula of the solid product.

If anyone could help me out with this, that would be great! Its 3 marks by the way
2. Sorry you've not had any responses about this. Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get more responses. Hopefully someone will be able to get back to you
3. SrCl2 . 6H2O ------> SrCl2 . XH2O + H20
0.02mol 0.08mol
0.08/0.02 = 4 as 1:4 ratio
6-4= 2
4. Could you explain the ratio please?
5. First work out mol of original : 3.892/266.6 = 0.02
Mol of water lost = (5.332-3.892) / 18 (Mr of H2O) = 0.08
This means that for 0.02 mol of SrCl2.6H2O that reacts, 0.08 mol of water is produced
You can either multiply both by 50, or just do 0.08/0.02 to work out for each mol of SrCl2.6H2O 4 mol of water is made.
The rest is easy, if 4 mol of water are lost, then 6-4=2 mol remain, so compound is:
SrCl2.2H2O

Hope this helps
6. (Original post by azo)
First work out mol of original : 3.892/266.6 = 0.02
Mol of water lost = (5.332-3.892) / 18 (Mr of H2O) = 0.08
This means that for 0.02 mol of SrCl2.6H2O that reacts, 0.08 mol of water is produced
You can either multiply both by 50, or just do 0.08/0.02 to work out for each mol of SrCl2.6H2O 4 mol of water is made.
The rest is easy, if 4 mol of water are lost, then 6-4=2 mol remain, so compound is:
SrCl2.2H2O

Hope this helps
Thanks, finally got my head around it!!

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Updated: March 31, 2015
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