Calculating % of Calcium Carbonate in Limestone

In summary, the conversation discusses the process of dissolving limestone in hydrochloric acid solution and neutralizing it with sodium hydroxide. The resulting solution requires calculations to determine the amount of calcium carbonate present. The steps involved include calculating the moles of HCl and NaOH, determining the amount of HCl originally used, and ultimately finding the moles of CaCO3 present.
  • #1
Trail_Builder
149
0
I thought I knew moles inside out, but I havnt quite been told how to factor in what happens when you dissolve something in something.

Homework Statement



1.500g of a sample of limestone was dissolved in 50cm^3 of a 1.00 mol/dm^3 of hydrochloric acid solution. The resulting solution was made up to 250cm^3 precisely with distilled water. 25.0cm^3 of this solution required 21.05cm^3 of 0.100 mol/dm^3 of sodium hydroxide for neutralisation.

Calculate the % of rock which is calcium carbonate.

Homework Equations




The Attempt at a Solution



mol of HCL:

mol = (vol/1000)*conc
50cm^3 is 1 mol/dm^3, so added with water up to 250cm^3 will be 0.2 mol/dm^3

mol(HCl) = 0.250*0.2
mol(HCl) = 0.05

but then we only take 25cm^3, so...

mol(HCl) = 0.005

mol of NaOH:

mol(NaOH) = (vol/1000)*conc
mol(NaOH) = (21.05/1000)*0.1

mol(NaOH) = 0.002105


Now I get stuff, I am not sure what the effect of dissolving 1.5g of limestone will even have, so I am stuck :S.


Hope you can help.


Thanks
 
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  • #2
1.500g of a sample of limestone was dissolved in 50cm^3 of a 1.00 mol/dm^3 of hydrochloric acid solution. The resulting solution was made up to 250cm^3 precisely with distilled water. 25.0cm^3 of this solution required 21.05cm^3 of 0.100 mol/dm^3 of sodium hydroxide for neutralisation.
The NaOH is needed to neutralise any HCl that wasn't 'used up' by the limestone - you just have to work backwards.

1, Work out many moles of NaOH
2, How many moles of HCl would this react with
3, How much HCl was made up originally.
4, - subtract (2)
5, how much CaCO3 would this HCl react with?
 
  • #3
so for 1) I have 0.002105
for 2) 0.002105 (because 1:1 ratio)
3) 0.005
4) 0.002895
5) 0.0014475 mol of CaC03 right? (because the equation would be CaCO3 + 2HCl goes to CaCl2 + H2O + CO2, so 1:2 ratio)

just checking lol.

thanks for your help :D
 

Related to Calculating % of Calcium Carbonate in Limestone

1. What is the formula for calculating the % of Calcium Carbonate in Limestone?

The formula for calculating the % of Calcium Carbonate in Limestone is: % CaCO3 = (grams of CaCO3 / grams of limestone) x 100

2. How do I determine the grams of CaCO3 in a sample of limestone?

To determine the grams of CaCO3 in a sample of limestone, you will need to first weigh the sample using a balance. Then, you will need to use a chemical reaction (such as adding hydrochloric acid) to dissolve the limestone and isolate the CaCO3. Finally, you will need to weigh the remaining CaCO3 and subtract it from the original weight of the sample to determine the grams of CaCO3.

3. Can I use any type of limestone for this calculation?

No, the type of limestone used for this calculation should be pure and free of any impurities. This means that it should not contain any other minerals or substances, as this will affect the accuracy of the calculation.

4. Why is it important to calculate the % of Calcium Carbonate in Limestone?

Calculating the % of Calcium Carbonate in Limestone is important because it is the main component of limestone and is responsible for its characteristic properties, such as its ability to neutralize acids. Knowing the % of CaCO3 in a sample of limestone can also provide valuable information for industries that use limestone as a raw material.

5. Are there any other methods for determining the % of Calcium Carbonate in Limestone?

Yes, there are other methods for determining the % of Calcium Carbonate in Limestone, such as using a spectrophotometer or titration techniques. However, the method of using a chemical reaction and measuring the resulting CaCO3 is the most commonly used and accurate method.

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