A) 0.36 g
B) 0.14 g
C) 0.28 g
D) 1.08 g
Correct Answer: C
Solution :
| [c] Idea. While solving this problem, students are advised to use the mole concept and stoichiometry as follows |
| Consider arbitrary mass of \[Mg{{C}_{2}}{{O}_{4}}\] and\[Ca{{C}_{2}}{{O}_{4}}(0.7-x)\] and x respectively |
| Write the chemical reaction and calculate weight of both species. |
| Finally using information provided in question complete the further calculation. |
| Let x g = weight of \[Ca{{C}_{2}}{{O}_{4}}\] |
| So, wt. of Mg \[{{C}_{2}}{{O}_{4}}=(0.7-x)g\] |
| \[Ca{{C}_{2}}{{O}_{4}}\xrightarrow{\Delta }CaC{{O}_{3}}+C{{O}_{2}}\] |
| \[Mg{{C}_{2}}{{O}_{4}}\xrightarrow{\Delta }MgC{{O}_{3}}+C{{O}_{2}}\] |
| Weight of \[CaC{{O}_{3}}\] produced \[=\frac{x}{128}\times 100\] |
| Weight of \[MgC{{O}_{3}}\] produced\[=\frac{0.7-x}{112}\times 84\] |
| \[\frac{x}{128}\times 100+\frac{0.7-x}{112}\times 84=0.47\] |
| \[x=0.46g\] |
| Mol. wt. \[CaC{{O}_{3}}=100,\]\[MgC{{O}_{3}}=84,\]\[Ca{{C}_{2}}{{O}_{4}}=128,\] \[Mg{{C}_{2}}{{O}_{4}}=112\] |
| Due to further heating |
| \[CaC{{O}_{3}}\xrightarrow{\Delta }\underset{\frac{x}{128}}{\mathop{CaO}}\,+C{{O}_{2}}\] |
| \[MgC{{O}_{3}}\xrightarrow{\Delta }\underset{\frac{7-x}{112}}{\mathop{MgO}}\,+C{{O}_{2}}\] |
| Weight of \[CaO\]and \[MgO\] |
| \[=\frac{0.46}{128}\times 56+\frac{0.24}{112}\times 40\] |
| \[=0.20+0.0857\] |
| = 0.28 g |
| TEST Edge. Students are advised to also go through clear idea about calculation of amount of product obtained in various chemical reaction which may be titration, precipitation, combustion reaction or any redox reaction using mole concept and stoichiometry. |
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