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:
$$CaC{O}_3+2HCl\stackrel{}{\to }CaC{l}_2+H_{2}O+C{O}_2$$
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 = $$[N\times E\times {\displaystyle \frac{V}{1000}}]$$
= $$[50\times {\displaystyle \frac{1}{10}}\times {\displaystyle \frac{150}{1000}}]$$
= 0.75 g
So, the percentage purity of calcium carbonate = total mass of pure $$CaC{O}_3$$/ total mass of impure solution x 100
       = $${\displaystyle \frac{0.75}{1}}\times 100$$
       = 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 $$HCl$$.