# Calcium Carbonate lab design

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I was wondering what is the best process to measure the CaCo3 (calcium carbonate) content of paper. I was thinking of making a reaction between CaCO3 with HCl. From that, how can I measure the CaCO3 (from the fact that it is also the limiting reagent)?

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I guess you could add excess HCl and measure the volume of CO2 released. Once the volume remains constant for about 2 minutes it means that all the calcium carbonate has been used up. With that volume, use the p x v = n R T equation to calculate the number of moles of CO2 which, in turn, is the same as the initial number of moles of CaCO3.

CaCO3 + 2HCl --------> CaCl2 + H2O + CO2 (Sorry for the subindices missing)

Weigh the paper before carrying out the experiment and then multiply the number of moles obtained by the molar mass of calcium carbonate and compare both. I think this might help you solve your problem.

Don't forget to repeat the test to minimise experimental errors.

Regards.

P.S. You might have to tear the paper into pieces to increase surface area and place the container with the reaction in a water bath to increase speed. (The problem with the water bath is that you won't know at what temperature will CO2 be - an intermediate temperature between the water's temperature and room temperature - so try it first without the water bath and if nothing happens do use it).

Edited by Hedron123

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I guess you could add excess HCl and measure the volume of CO2 released. Once the volume remains constant for about 2 minutes it means that all the calcium carbonate has been used up. With that volume, use the p x v = n R T equation to calculate the number of moles of CO2 which, in turn, is the same as the initial number of moles of CaCO3.

CaCO3 + 2HCl --------> CaCl2 + H2O + CO2 (Sorry for the subindices missing)

Weigh the paper before carrying out the experiment and then multiply the number of moles obtained by the molar mass of calcium carbonate and compare both. I think this might help you solve your problem.

Don't forget to repeat the test to minimise experimental errors.

Regards.

P.S. You might have to tear the paper into pieces to increase surface area and place the container with the reaction on a water bath to increase speed. (The problem with the water bath is that you won't know at what temperature will CO2 be - an intermediate temperature between the water's temperature and room temperature - so try it first without the water bath and if nothing happens do use it).

Couldn't you just use the volume of water it produced? do some simple conversions and its a 1:1 ratio just like CO2 so its just a 1 step conversion from volume to moles.

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I guess you could add excess HCl and measure the volume of CO2 released. Once the volume remains constant for about 2 minutes it means that all the calcium carbonate has been used up. With that volume, use the p x v = n R T equation to calculate the number of moles of CO2 which, in turn, is the same as the initial number of moles of CaCO3.

CaCO3 + 2HCl --------> CaCl2 + H2O + CO2 (Sorry for the subindices missing)

Weigh the paper before carrying out the experiment and then multiply the number of moles obtained by the molar mass of calcium carbonate and compare both. I think this might help you solve your problem.

Don't forget to repeat the test to minimise experimental errors.

Regards.

P.S. You might have to tear the paper into pieces to increase surface area and place the container with the reaction on a water bath to increase speed. (The problem with the water bath is that you won't know at what temperature will CO2 be - an intermediate temperature between the water's temperature and room temperature - so try it first without the water bath and if nothing happens do use it).

Couldn't you just use the volume of water it produced? do some simple conversions and its a 1:1 ratio just like CO2 so its just a 1 step conversion from volume to moles.

The fact is you will use a solution of HCl and with the paper on the bottom it will be harder to measure accurate volumes changes. I believe the CO2 production is a more feasible method but I'm open to other suggestions. Besides you cannot predict how much will the volume increase, maybe it's such an insignificant amount that it's hard to account.

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I guess you could add excess HCl and measure the volume of CO2 released. Once the volume remains constant for about 2 minutes it means that all the calcium carbonate has been used up. With that volume, use the p x v = n R T equation to calculate the number of moles of CO2 which, in turn, is the same as the initial number of moles of CaCO3.

CaCO3 + 2HCl --------> CaCl2 + H2O + CO2 (Sorry for the subindices missing)

Weigh the paper before carrying out the experiment and then multiply the number of moles obtained by the molar mass of calcium carbonate and compare both. I think this might help you solve your problem.

Don't forget to repeat the test to minimise experimental errors.

Regards.

P.S. You might have to tear the paper into pieces to increase surface area and place the container with the reaction on a water bath to increase speed. (The problem with the water bath is that you won't know at what temperature will CO2 be - an intermediate temperature between the water's temperature and room temperature - so try it first without the water bath and if nothing happens do use it).

Couldn't you just use the volume of water it produced? do some simple conversions and its a 1:1 ratio just like CO2 so its just a 1 step conversion from volume to moles.

The fact is you will use a solution of HCl and with the paper on the bottom it will be harder to measure accurate volumes changes. I believe the CO2 production is a more feasible method but I'm open to other suggestions. Besides you cannot predict how much will the volume increase, maybe it's such an insignificant amount that it's hard to account.

I see the whole harder to measure thing, I just never like trusting probes =/ Last time i used a pH probe it told me DISTILLED WATER's pH was about 2....

You could also find the mass of the water produced with heating your product which will be a solution. Mass the product before heating, mass it after evaporating as much of the water as possible, find the difference. Also since the mole ratio is 2HCl to 1 water in theory you should get half the mass of the HCl back as water.

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