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 \[HCl\].