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If 125g of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain?

Any help would be greatly appreciated!

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- Thread starter Raza
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- #1

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If 125g of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain?

Any help would be greatly appreciated!

- #2

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Do you know the formula for the above mixture ? What does the formula tell you about how much each part contributes to the mass ?Raza said:

If 125g of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain?

Any help would be greatly appreciated!

- #3

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is it Mg_{2}SO_{3} **.** 7H_{2}O?

I could probably solve this if I knew what anhydrous meant.

I could probably solve this if I knew what anhydrous meant.

- #4

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Anhydruous=without water...

Daniel.

- #5

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I am sorry, I do not get this; could you please rephrase it for me?What's sulfyte?

and what is a difference between dehyration and anhydrate?

and what is a difference between dehyration and anhydrate?

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- #6

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Raza said:I am sorry, I do not get this; could you please rephrase it for me?What's sulfyte?

and what is a difference between dehyration and anhydrate?

[itex]MgSO_{3} [/itex] is MAGNESIUM SULPHYTE and [itex] MgSO_{4} [/itex] is MAGNESIUM SULPHATE.One of them comes from the sulphurous acid,the other from sulphuric acid...

DEHYDRATION=losing water.

ANHYDROUS=without water...

Daniel.

- #7

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I finally get the question but how do I solve it?

I done this before but with knowing the weight of each elements.

I done this before but with knowing the weight of each elements.

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- #8

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Daniel.

- #9

GCT

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You know that for every one mole of the compound, there is 7 moles of H20. This should give you a stoichiometric ratio 7 moles of H20/1 mole of the compound. Convert using this ratio to moles of water. From this mole value, convert to grams of water. Subtract this value from 125, and that should be your answer.

- #10

saltydog

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I'd like to help:

mol. wgt of MgS04*7H20 = 246.3

mol.wgt of MgS04 = 120.3

Thus, the ratio of anhydrous/hydrous (without water/with water) = 0.488

Thus, when you remove the water from 125 grams of the hydrate, the remaining is:

125*0.488=61 grams MgSO4 left.

I believe that's right. It's been a while and I don't want to make it worst for you.

Salty

mol. wgt of MgS04*7H20 = 246.3

mol.wgt of MgS04 = 120.3

Thus, the ratio of anhydrous/hydrous (without water/with water) = 0.488

Thus, when you remove the water from 125 grams of the hydrate, the remaining is:

125*0.488=61 grams MgSO4 left.

I believe that's right. It's been a while and I don't want to make it worst for you.

Salty

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- #11

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Thank you all, I got how you solve it now.

It seems so easy after you get what is doing on

It seems so easy after you get what is doing on

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