A mixture of $\mathrm{MgCO}_{3}$ and $\mathrm{CaCO}_{3}$ was heated for a long time, the weight is d

Question

A mixture of $\mathrm{MgCO}_{3}$ and $\mathrm{CaCO}_{3}$ was heated for a long time, the weight is decreased by $50 \%$. Then the correct statement about percentage composition of $\mathrm{MgCO}_{3}$ and that of $\mathrm{CaCO}_{3}$ in the mixture is/are\ I. Percentage of $\mathrm{MgCO}_{3}$ in sample is $71.5 \%$.\ II. Percentage of $\mathrm{CaCO}_{3}$ in sample is $28.5 \%$.\ III. Percentage of $\mathrm{MgCO}_{3}$ in sample is $84 \%$.\ IV. Percentage of $\mathrm{CaCO}_{3}$ in sample is $16 \%$.\

Professor PC Thomas Classes · Answer

(c) : The given equations are as follows:\ $\mathrm{MgCO}_{3(s)} \xrightarrow{\Delta} \mathrm{MgO}_{(s)}+\mathrm{CO}_{2(g)}$\ $\mathrm{CaCO}_{3(s)} \xrightarrow{\Delta} \mathrm{CaO}_{(s)}+\mathrm{CO}_{2(g)}$\ Let total mass of the mixture $=1.0 \mathrm{~g}$\ Then mass of $\mathrm{MgCO}_{3}=x \mathrm{~g}$ and mass of $\mathrm{CaCO}_{3}=(1-x) \mathrm{g}$\ According to the above equation,\ $84 \mathrm{~g} \mathrm{MgCO}_{3}$ gives $44 \mathrm{~g} \mathrm{CO}_{2}$\ $	herefore \quad x \mathrm{~g} \mathrm{MgCO}_{3}$ gives $\frac{44}{84} x \mathrm{~g} \mathrm{CO}_{2}$\ Again $100 \mathrm{~g} \mathrm{CaCO}_{3}$ gives $44 \mathrm{~g} \mathrm{CO}_{2}$\ $	herefore \quad(1-x) \mathrm{g} \mathrm{CaCO}_{3}$ gives $\frac{44}{100}(1-x) \mathrm{g} \mathrm{CO}_{2}$\ According to the problem, $\frac{44 x}{84}+\frac{44}{100}(1-x)=0.5 \mathrm{~g}$\ On solving we get, $x=0.715 \mathrm{~g}$\ Hence, percentage of $\mathrm{MgCO}_{3}=0.715 	imes 100=71.5 \%$ and percentage of $\mathrm{CaCO}_{3}=100-71.5=28.5 \%$