Question

150 mL of N/10 $HCl$ is required to react completely with 1.0 g of a sample of limestone. Calculate the percentage purity of calcium carbonate.

Answer 1

Hint: We know that purity of a substance is the measure of the extent to which a given substance is pure. The percentage purity has to be used in stoichiometric calculations. And limestone is $CaC{{O}_{3}}$ which has a molar mass of 100.

Step By step solution:
We know that the percentage purity of a substance is the mass of a pure compound in the impure sample /total mass of impure sample x100. And the reaction of $HCl$ and limestone is:
$CaC{{O}_{3}}+2HCl\xrightarrow{{}}CaC{{l}_{2}}+{{H}_{2}}O+C{{O}_{2}}$
In the problem it is given that a given amount of $HCl$ is required to completely react with limestone.
So, 150 mL of N/10 HCl$$$$\equiv 150 mL of N/10 $CaC{{O}_{3}}$
And equivalent mass of $CaC{{O}_{3}}$ = Total molecular mass of $CaC{{O}_{3}}$/ number of equivalent
= (40+12+48)/2
= 100/2
= 50 g/eq.

So, total mass of $CaC{{O}_{3}}$ present in 150 mL of N/10 solution = $[N\times E\times \dfrac{V}{1000}]$
= $[50\times \dfrac{1}{10}\times \dfrac{150}{1000}]$
= 0.75 g
So, the percentage purity of calcium carbonate = total mass of pure $CaC{{O}_{3}}$/ total mass of impure solution x 100
= $\dfrac{0.75}{1}\times 100$
= 75%
So, the answer is 75%.

Note: We should remember that for calculating % purity we use equivalent weight not molecular weight. We have to calculate the number of equivalents from the balanced chemical reaction of $CaC{{O}_{3}}$ and $HCl$.