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 $O{H^ - }$ .