Question

A sample of calcium carbonate is $80\%$ pure, $25$ gm of this sample is treated with excess of $HCl$. How much volume of $C{O}_2$ will be obtained at $1$ atm & $273$ K?

Answer 1

The mass of pure ${CaCO}_3=\frac{80}{100}\times 25=20g$

${CaCO}_3\left(s\right)+2HCl\left(aq\right)\to Ca{Cl}_2\left(s\right)+{CO}_2\left(g\right)+H_{2}O\left(l\right)$

The mole ratio of calcium carbonate and ${CO}_2$ is $1:1$

Moles of ${CaCO}_3=\frac{Massof{CaCO}_3}{Molarmass}=\frac{20}{100}=0.2$ moles

Moles of ${CO}_2=0.2$ moles

Molar gas volume $1$ mole $=22.4$ litres

$0.2$ mole $=2\times 22.4=4.48$ litres.

$4.48$ litres of ${CO}_2$ will be obtained.