JAMB Chemistry 2019 Paper

Concept:Heating calcium hydrogen trioxocarbonate(IV) causes thermal decomposition, producing a calcium carbonate precipitate.Explanation:The decomposition equation is:$$\mathrm{Ca(HCO_3)_2 \xrightarrow{\Delta} CaCO_3(s) + H_2O(g) + CO_2(g)}$$One mole of $\mathrm{Ca(HCO_3)_2}$ gives one mole of $\mathrm{CaCO_3}$ precipitate.First, convert the volume to $\mathrm{dm^3}$:$$200\ \mathrm{cm^3} = 0.2\ \mathrm{dm^3}$$Calculate the moles of calcium hydrogen trioxocarbonate(IV):$$\text{Moles} = \text{Concentration} \times \text{Volume} = 0.50 \times 0.2 = 0.1\ \mathrm{mol}$$Molar mass of $\mathrm{CaCO_3}$:$$40 + 12 + (3 \times 16) = 100\ \mathrm{g/mol}$$Since $1\ \mathrm{mol}$ of $\mathrm{Ca(HCO_3)_2}$ yields $100\ \mathrm{g}$ of $\mathrm{CaCO_3}$:$$0.1\ \mathrm{mol} \times 100\ \mathrm{g/mol} = 10\ \mathrm{g}$$Answer:The maximum weight of solid precipitated is $10\ \mathrm{g}$, which is option A.