• # question_answer A 0.60 g sample consisting of only ${{\operatorname{CaC}}_{2}}{{O}_{4}}$and $Mg{{C}_{2}}{{O}_{4}}$ is heated at $500{}^\circ C$, converting the two salts to ${{\operatorname{CaCO}}_{3}}$, and $MgC{{O}_{3}}$. The sample then weighs 0.465 g. If the sample had been heated to$900{}^\circ C$, where the products are $CaO$ and $MgO$. What would the mixtures of oxides have weighed? A) 0.12g B) 0.21 g C) 0.252g D) 0.3 g

 Let $x$g is mass of ${{\operatorname{CaC}}_{2}}{{O}_{4}}$and$\left( 0.6-x \right)$g mass of ${{\operatorname{MgC}}_{2}}{{O}_{4}}.$ ${{\operatorname{CaC}}_{2}}{{O}_{4}}\xrightarrow{\Delta }CaC{{O}_{3}}+C{{O}_{2}}$ ${{\operatorname{MgC}}_{2}}{{O}_{4}}\xrightarrow{\Delta }{{\operatorname{MgCO}}_{3}}+C{{O}_{2}}$ Mass of ${{\operatorname{CaCO}}_{3}}$ produced $=\frac{x}{128}\times 100$
 Mass of ${{\operatorname{MgCO}}_{3}}$produced $=\frac{\left( 0.6-x \right)}{112}\times 84$ $\therefore \,\,\,\frac{x}{128}\times 100+\frac{\left( 0.6-x \right)}{112}\times 84=0.465$ $x=0.48g$ On further heating, ${{\operatorname{CaCO}}_{3}}\xrightarrow{\Delta }\operatorname{CaO}+C{{O}_{2}}$ ${{\operatorname{MgCO}}_{3}}\xrightarrow{\Delta }\overset{\frac{x}{128}}{\mathop{\underset{\left( \frac{0.6-x}{112} \right)}{\mathop{\operatorname{Mg}}}\,}}\,O+C{{O}_{2}}$ Mass of CaO b=and MgO produced$=\frac{0.48}{128}\times 56+\frac{0.12}{112}\times 40=0.252\operatorname{g}$