[Monoceros]

Hello,

I am new to this list and subscribed because I have no clue how to resolve this reaction :

F6 H8 N2 Si  +  Ca (OH2) -> ?

Can anyone help me out ?

Thank you.

[Borek]

I am new to this list

This is not a list, this is forum

I have no clue how to resolve this reaction :

F6 H8 N2 Si  +  Ca (OH2) -> ?

Start deciphering what F₆H₈N₂Si is.

[Monoceros]

Oh ... I'm sorry.

Can you tell me where I would get an answer ? 

I should calculate the amount of chalk needed for the ammonium hexafluorosilicate but I don't understand the reaction ... my chemical lessons unfortunately date back 25 years ago.

[Arkcon]

Well, your given formula doesn't match the name ammonium hexafluorosilicate, so it's a little hard to help.  Perhaps a technical reference for your given field has a table, or fudge-factor calculation, that you can google for by name?  Can you give us some more info, what do you do, what reaction were you expecting, etc.

[Monoceros]

The formula of the product is (NH4)2SiF6.

It is a waste water : water with the product after a cleaning operation.

We want to add chalk to flocculate the F.

[Arkcon]

Well, if the fluorine in the molecule ionizes completely in solution, it will form the insoluble product CaF₂.  This is a common way to remediate fluorine.  Maybe you'd like to try and write a balanced reaction for this, yourself?  And maybe you can figure out how to convert moles to grams, to determine the amount needed?  The more you know about this subject, the more useful you'll be, and better able to handle your next problem.

[Monoceros]

I arrive at this :

(NH4)2SiF6 + 3 Ca(OH)2 -> 3 CaF2 + SiO2 + 2 NH3 + 3 H20

If this is correct, then :
1 mol (NH4)2SiF6 needs 3 mol chalk
1 mol (NH4)2SiF6 means 178,2 g (of which 18,99g is F)

We have an analysis of 8 mg/l F.

So this would mean on a delivery of 30.000 liter : 240 g F which would mean that we have 240/18,99 = 12,6 mol (NH4)2SiF6.

Therefore we need 3 x 12,6 = 37,9 mol of chalk or 37,9 x 74,1 = 2809 g of chalk.

I hope this is the right way to calculate.

[Borek]

240 g F which would mean that we have 240/18,99 = 12,6 mol (NH4)2SiF6.

No, there is 6*18,99g F per mole.

[AWK]

1 mol (NH4)2SiF6 needs 3 mol chalk

This is not a chalk, this is a lime

[Monoceros]

Yes, of course ... sorry for my bad english 

[Monoceros]

240 g F which would mean that we have 240/18,99 = 12,6 mol (NH4)2SiF6.

No, there is 6*18,99g F per mole.

OK, I understand.  Then it would be :

(NH4)2SiF6 + 3 Ca(OH)2 -> 3 CaF2 + SiO2 + 2 NH3 + 3 H20

If this is correct, then :
1 mol (NH4)2SiF6 needs 3 mol lime
1 mol (NH4)2SiF6 means 178,2 g (of which 113,94g is F)

We have an analysis of 8 mg/l F.

So this would mean on a delivery of 30.000 liter : 240 g F which would mean that we have 240/113,94 = 2,1 mol (NH4)2SiF6.

Therefore we need 3 x 2,1 = 6,3 mol of lime or 6,3 x 74,1 = 468 g of lime.

I hope I got it right now.

[Borek]

Therefore we need 3 x 2,1 = 6,3 mol of lime or 6,3 x 74,1 = 468 g of lime.

Stochiometry is OK

But I am not sure if it will work this way.

Your solution has concentration of about 7x10^-5M. Assuming it will hydrolise completely to HF (unlikely)  that gives 4.2x10^-4M F^-. At this concentration HF will be dissociated in about 60%, giving 2.6x10^-4M F^-. pKso of CaF₂ is 10.5 so you will ned concentration of Ca²⁺ in the range of 10^-10.5/(2.6x10^-4)² = 4.7x10^-4 to start precipitation. That means at least 1050 g of Ca(OH)₂. That's not whole truth, as adding strong base will move pH up, increasing dissociation of HF, thus decreasing minimum concentration of Ca²⁺, but I am estimating here only start of the precipitation, and even that with not so realistic assumption about hydrolysis.

[Monoceros]

OK.

Thank you all for your input.