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Topic: Crystalohydrates  (Read 6100 times)

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Offline justukyte

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Crystalohydrates
« on: January 29, 2012, 06:43:22 AM »
1) We had 500 g of MgSO4*7H2O, after leaving it an open flask for a while its mass decreased to 390 g. How much crystalized water is left in the crystals? What is the formula of the newly formed crystalohydrate?

How to solve this problem? Can anyone explain to me what happened with the crystalohydrate? What kind of reaction should I write, is it even necessary?

2) Another problem: What volume of 3M H2SO4 solution reacted with Fe, if the outcome was 55,6 g of FeSO4*7H2O?

Fe + H2SO4 -> FeSO4*7H2O. Is this kind of reaction possible? If no, than how should I get the answer to this problem?

I'm a non-English student, sorry for my bad English.  ::)
« Last Edit: January 29, 2012, 06:53:51 AM by justukyte »

Offline UG

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Re: Crystalohydrates
« Reply #1 on: January 29, 2012, 06:50:04 AM »
The mass decreased because water has been lost. Can you work out how many moles of MgSO4.7H2O there is in 500 g? Also, can you work out how many moles of water was lost?

Offline Borek

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Re: Crystalohydrates
« Reply #2 on: January 29, 2012, 06:52:47 AM »
You can't write an exact reaction equation (yet), as you don't know (yet) what the product is.

What is happening is that your crystalline water is leaving the crystals, leaving some mixture of different hydrates. It can be expressed with an equation looking like:

MgSO4·7H2O -> MgSO4·(7-x)H2O + xH2O

Then it is a simple stoichiometry - use masses given to find x.

Edit: UG hinted at exactly the same approach.
« Last Edit: January 29, 2012, 08:45:04 AM by Borek »
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Offline justukyte

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Re: Crystalohydrates
« Reply #3 on: January 29, 2012, 06:59:26 AM »
Thanks, UG and Borek, I think I got the idea how to solve the first problem.  :)
You are very quick to reply, when I was  about to write another question about crystalohydrates, you had already answered.

Anyway, thanks :)

Offline UG

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Re: Crystalohydrates
« Reply #4 on: January 29, 2012, 07:06:39 AM »
2) Another problem: What volume of 3M H2SO4 solution reacted with Fe, if the outcome was 55,6 g of FeSO4*7H2O?
How many moles of FeSO4.7H2O is there in 55.6 g? This is the number of moles of SO42- present in the crystalohydrate, and this will come from the H2SO4 solution. Do you know the relationship between volume, number of moles and concentration?

Offline justukyte

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Re: Crystalohydrates
« Reply #5 on: January 29, 2012, 07:14:12 AM »
2) Another problem: What volume of 3M H2SO4 solution reacted with Fe, if the outcome was 55,6 g of FeSO4*7H2O?
How many moles of FeSO4.7H2O is there in 55.6 g? This is the number of moles of SO42- present in the crystalohydrate, and this will come from the H2SO4 solution. Do you know the relationship between volume, number of moles and concentration?

Yeah, I know that 3M means 3 mol/l and the formula (C=n/v), but I wasn't sure how to begin solving this problem. Isn't there any other product from the reaction other than the crystalohydrate, for example, SO2, SO3 or anything else with sulphur or iron? Just the plain crystalohydrate?

Offline UG

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Re: Crystalohydrates
« Reply #6 on: January 29, 2012, 07:17:34 AM »
From the information given in the question, I think the only product would just be the crystalohydrate.

Offline justukyte

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Re: Crystalohydrates
« Reply #7 on: January 29, 2012, 07:19:19 AM »
Ok, then if there's no other product, I think I'll be able to solve it. Thanks again!  :)

Offline Borek

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Re: Crystalohydrates
« Reply #8 on: January 29, 2012, 08:44:40 AM »
Fe + H2SO4 -> FeSO4*7H2O

Actually there is another product, also, there is another reactant. Full, unbalanced reaction equation is:

Fe + H2SO4 + H2O -> FeSO4·7H2O + H2

However, the only thing that matters here is the molar ratio between iron and sulfuric acid, and none of the other substances present is going to change it.
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