JAMB Chemistry 2020 Paper

Concept:The number of product molecules is found using stoichiometry, the mole concept, and Avogadro's constant.Explanation:Calculate the moles of $\mathrm{CaCO}_3$ used.Molar mass of $\mathrm{CaCO}_3$ is $100\ \mathrm{g\,mol^{-1}}$.Amount of $\mathrm{CaCO}_3 = \frac{10.0\ \mathrm{g}}{100\ \mathrm{g\,mol^{-1}}} = 0.10\ \mathrm{mol}$.Calculate the moles of $\mathrm{HCl}$ supplied.Amount of $\mathrm{HCl} = 0.2\ \mathrm{dm^3} \times 1\ \mathrm{mol\,dm^{-3}} = 0.20\ \mathrm{mol}$.The balanced equation is:$\mathrm{CaCO}_3 + 2\mathrm{HCl} \rightarrow \mathrm{CaCl}_2 + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}$.From the equation, $1\ \mathrm{mol}$ of $\mathrm{CaCO}_3$ requires $2\ \mathrm{mol}$ of $\mathrm{HCl}$.So $0.10\ \mathrm{mol}$ of $\mathrm{CaCO}_3$ requires $0.20\ \mathrm{mol}$ of $\mathrm{HCl}$, which is exactly available.Therefore, $0.10\ \mathrm{mol}$ of $\mathrm{CO}_2$ is produced.Number of molecules of $\mathrm{CO}_2$:$0.10 \times 6.02 \times 10^{23} = 6.02 \times 10^{22}$ molecules.Answer:$6.02 \times 10^{22}$ molecules of Carbon(iv)Oxide are produced.Correct option: C.