What mass of an $80\%$ pure magnesium carbonate ($\mathrm{MgCO}_3$) sample will be required to compl

Question

What mass of an $80\%$ pure magnesium carbonate ($\mathrm{MgCO}_3$) sample will be required to completely neutralize $100 \mathrm{~mL}$ of $0.5 \mathrm{~M} \mathrm{HCl}$ solution? (Molar mass of $\mathrm{MgCO}_3 = 84 \mathrm{~g/mol}$)

Professor PC Thomas Classes · Answer

(b) : The neutralization reaction is $\mathrm{MgCO}_3 + 2\mathrm{HCl} ightarrow \mathrm{MgCl}_2 + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}$. The number of moles of $\mathrm{HCl} = 	ext{Molarity} 	imes 	ext{Volume in Liters} = 0.5 	imes 0.100 = 0.05 \mathrm{~mol}$. From the stoichiometry, 2 moles of $\mathrm{HCl}$ require 1 mole of $\mathrm{MgCO}_3$. Thus, the required moles of pure $\mathrm{MgCO}_3 = 0.05 / 2 = 0.025 \mathrm{~mol}$. The mass of pure $\mathrm{MgCO}_3$ required $= 0.025 	imes 84 = 2.10 \mathrm{~g}$. Since the sample is only $80\%$ pure, the mass of the impure sample required is $\frac{2.10}{0.80} = 2.625 \mathrm{~g}$.