A) 0.48 g
B) 0.80 g
C) 0.42 g
D) 0.28 g
Correct Answer: A
Solution :
\[\underset{\begin{smallmatrix} 1\,mol \\ \frac{0.42}{84}=0.005 \end{smallmatrix}}{\mathop{MgC{{O}_{3}}}}\,\xrightarrow{\Delta }\underset{\begin{smallmatrix} 1\,mol \\ 0.005\,mol \end{smallmatrix}}{\mathop{MgO}}\,+C{{O}_{2}}\] \[\underset{\begin{smallmatrix} 1\,mol \\ \frac{0.5}{100}=0.005 \end{smallmatrix}}{\mathop{CaC{{O}_{3}}}}\,\xrightarrow{\Delta }\underset{\begin{smallmatrix} 1\,mol \\ 0.005\,mol \end{smallmatrix}}{\mathop{CaO}}\,+C{{O}_{2}}\] Total weight \[=wt.\,of\,MgO+wt\,of\,CaO\] \[=0.005\times 40+0.005\times 56\] \[=0.48g\]
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