llennon_04
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Last week my class partook in a chemistry practical as part of our coursework. The practical was to make a salt (magnesium sulfate crystals). I am rather apprehensive over a few of the following questions, and so I am addressing all chemists on TSR for help.

1. "Suggest a reason why the procedure in step 2 and 3 is only possible with a carbonate that is insoluble in water such as magnesium carbonate."
>Step 2 was adding magnesium carbonate to sulfuric acid and stirring until the solution went clear, and repeating this step until the solution would no longer clear.
>Step 3 was to heat the beaker containing the solution until it became clear again, and then to add more magnesium carbonate so that some solid was in excess.
Was this something pertaining neutralisation reactions or?

2. "Suggest how your sample of magnesium sulfate would be different if in step 6, all of the water had been evaporated off."
>Step 6 being to boil the solution of magnesium sulfate on a tripod gauze until about 2/3 water evaporates.
Presumably wanting you to discuss the state of the crystals after total evaporation(?)

3. "Suggest one reason why a student carrying out the preparation apparently obtains a yield over 100%."
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TSR Jessica
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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
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ahmed.m
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can someone answer this please
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Asurat
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Answer to 2 is to do with Thermal decomposition I think, group 2 carbonates decompose with enough heat. Since there was water the water evaporated rather than decomposed
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Asurat
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In the first part is because if the solution goes clear the first time it's because of the reaction, and then it saturates afterwards due to the excess.

This would not be the case if the carbonate was soluble in water, as first time round there would be the reaction and second time round the carbonate would just dissolve in the water, giving a false-negative.
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