A) 0.12g
B) 0.21 g
C) 0.252g
D) 0.3 g
Correct Answer: C
Solution :
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\[=\frac{0.48}{128}\times 56+\frac{0.12}{112}\times 40=0.252\operatorname{g}\] |
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