Question

$50$ g of calcium carbonate sample, $CaC{O}_3$ (lime stone) gave $1.12$ L carbon dioxide gas measured at STP ($1$ mole of every ideal gas at STP has $22.4$ L volume).
What is percent purity of $CaC{O}_3$?

Answer 1

$50$ g of $CaC{O}_3$ corresponds to $\frac{50}{100}=0.5$ moles.
$1.12$ L of $C{O}_2$ corresponds to $\frac{1.12}{22.4}=0.05$ moles of $C{O}_2$.
Hence, percent purity of $CaC{O}_3=\frac{0.05}{0.5}\times 100=10\%$.