Two part solutions

Samandar · Jul 24, 2021

Hi Mr Holmes
I have followed you since 2010.
Newly I have calculated the necessity of cacl2 and Na2CO3 for my new tank. I wrote the following formula
CaCl2(aq) +Na2CO3 (aq) <---> CaCO3(s) +2NaCl(aq)
So as you see, for each mol of cacl2 we need only one mol Na2CO3. On the other hand for 147 gr cacl2 we need only 106 gr Na2CO3. Is it right? Or I have missed something.

CaCl2 is dihydrate.
(CaCl2-2H2O) with mol weight 147 gr.
(Na2Co3) with mol weight 106 gr.

P-Dub · Jul 24, 2021

Samandar said:
Hi Mr Holmes
I have followed you since 2010.
Newly I have calculated the necessity of cacl2 and Na2CO3 for my new tank. I wrote the following formula
CaCl2(aq) +Na2CO3 (aq) <---> CaCO3(s) +2NaCl(aq)
So as you see, for each mol of cacl2 we need only one mol Na2CO3. On the other hand for 147 gr cacl2 we need only 106 gr Na2CO3. Is it right? Or I have missed something.

CaCl2 is dihydrate.
(CaCl2-2H2O) with mol weight 147 gr.
(Na2Co3) with mol weight 106 gr.

@Randy Holmes-Farley

Randy Holmes-Farley · Jul 26, 2021

The equation and values are are basically right, but the consumption of calcium is a little less because some magnesium and strontium get into the deposited calcium carbonate in place of some of of the calcium.

This has more on DIY two part systems:

An Improved Do-it-Yourself Two-Part Calcium and Alkalinity Supplement System by Randy Holmes-Farley - Reefkeeping.com

Samandar · Jul 28, 2021

I have reviewed your paper. In reclpe #1 there is mention that for 500 gr CaCl2 dehydrate we need 594 gr baking soda (NaHCO3). Also we should heated backing soda at 300°F for one hour.
So, after heating we will have soda ash.
It means that we need for every mol of CaCl2-2H2O only one mol of Na2CO3. Another word, for every 147 gr CaCl2 - H2O, only 106 gr Na2CO3 is necessary.
Or we can say, for 500 gr CaCl2-2H2O we need 360 gr Na2CO3.
But 594 gr baking soda is equal to 375 gr soda ash. It means we used some Na2CO3 more than calculated.
Am I right?
What is the couse?

Randy Holmes-Farley · Jul 28, 2021

I'm not sure I understand your concerns. Your equations and numbers are accurate for making pure calcium carbonate.

500 grams of calcium chloride dihydrate (mw 147 g/mole) is 3.4 moles
594 grams of NaHCO3 (mw 84 g/mole) is 7.07 moles

If you fully bake 7.07 moles of NaHCO3, you get 3.54 moles of Na2CO3.

Thus, the calcium and carbonate are very close on a mole basis.

In reality, you want a little less calcium than the carbonate because some calcium gets replaced in the calcium carbonate structure by magnesium and strontium. Hence the mole ratio of consumption is a bit less calcium than carbonate.

Does that clarify things?

Samandar · Jul 30, 2021

Randy Holmes-Farley said:
I'm not sure I understand your concerns. Your equations and numbers are accurate for making pure calcium carbonate.

500 grams of calcium chloride dihydrate (mw 147 g/mole) is 3.4 moles
594 grams of NaHCO3 (mw 84 g/mole) is 7.07 moles

If you fully bake 7.07 moles of NaHCO3, you get 3.54 moles of Na2CO3.

Thus, the calcium and carbonate are very close on a mole basis.

In reality, you want a little less calcium than the carbonate because some calcium gets replaced in the calcium carbonate structure by magnesium and strontium. Hence the mole ratio of consumption is a bit less calcium than carbonate.

Does that clarify things?

Thank you for clear explanation.