Question

A $ 0.125g\; $ of a sample of limestone was dissolved in $ 30cc $ of $ \dfrac{N}{{10}}\;HCl\; $ and the solution diluted to $ 100cc $ . $ 10cc $ of this solution required $ 15\,cc $ of $ \dfrac{N}{{20}}\;NaOH\; $ for neutralization. Calculate $ \% $ of $ CaC{O_3} $ in limestone.

Answer 1

Hint : Limestone is the common name of the chemical compound calcium carbonate. It exists in rocks as the mineral calcite and aragonite. To calculate the $ \% $ of $ CaC{O_3} $ ​ in limestone, first we will calculate the moles of $ {H_2}C{O_3} $ and then we will equate it with the number of moles of $ NaOH $ used in the reaction.
Formula Used
Number of moles of a compound is given by the formula:
 $ No.\,of\,moles = \dfrac{{given\,mass}}{{molar\,mass}} $
Number of moles of a compound using normality and volume of the solution is given by the formula:
 $ No.\,of\,moles = Normality \times Volume $
The percentage composition of a substance is given by the formula:
 $ \% \,composition = \dfrac{{given\,mass}}{{total\,mass}} \times 100 $ .

Complete step by step solution
When limestone is treated with Hydrochloric acid it will form calcium chloride and hydrogen carbonate, the equation involved in the reaction will be:
 $ CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}C{O_3} $
Now, $ NaOH $ is used in the neutralization of hydrogen carbonate the equation involved in the reaction will be:
 $ {H_2}C{O_3} + 2NaOH \to N{a_2}C{O_3} + 2{H_2}O $
Given, amount of limestone dissolved in $ 30mL $ solution $ = 0.125g $
Normality of the $ HCl $ solution $ = \dfrac{1}{{10}} $
Normality of the $ NaOH $ solution $ = \dfrac{1}{{20}} $
Let the percentage of $ CaC{O_3} $ in the sample be $ x\,\% $ , so in a sample of $ 0.125g $ the amount of $ CaC{O_3} $ will be:
 $ amount\,of\,CaC{O_3} = \dfrac{{0.125x}}{{100}} $
Now, we will calculate the number of moles of calcium carbonate. It will be:
 $ No.\,of\,moles\,of\,CaC{O_3} = \dfrac{{given\,mass}}{{molar\,mass}} $
 $ No.\,of\,moles\,of\,CaC{O_3} = \dfrac{{0.125x}}{{100 \times 100}} = 0.125x \times {10^{ - 4}} $
After reaction calcium carbonate is converted into hydrogen carbonate so the number of moles in the solution will be equal, so;
 $ \therefore \,\,\,moles\,of\,{H_2}C{O_3} = moles\,of\,CaC{O_3} $
If we take $ 10\,mL $ of this solution from $ 100\,mL $ solution, then number of moles in $ 10\,mL $ solution will be;
 $ moles\,of\,{H_2}C{O_3} = 0.125x \times {10^{ - 4}} \times 10 = 0.125x \times {10^{ - 3}} $
Now, we will calculate the number of moles of $ NaOH $ used in the neutralization, it will be calculated as:
 $ moles\,of\,NaOH = Vol \times Normality\,of\,NaOH $
 $ moles\,of\,NaOH = \dfrac{{15}}{{{{10}^3}}} \times \dfrac{1}{{20}} $
Now from the equation we have,
 $ No.\,of\,moles\,of\,{H_2}C{O_3} = \dfrac{1}{2} \times No.\,of\,moles\,of\,NaOH\, $
 $ 0.125x\, \times {10^{ - 3}} = \dfrac{1}{2} \times \dfrac{{15}}{{2 \times {{10}^4}}}\, $
 $ x = \dfrac{{15 \times 10000}}{{5 \times 10000}} = 3\% $

Hence, the amount of calcium carbonate in the sample is $ 3\% $ .

Note
A neutralization reaction is defined as a chemical reaction between an acid and base quantitatively that react together to give salt and water as the products of the reaction. In this reaction there is a combination of $ {H^ + } $ or $ O{H^ - } $ .