A) 0.1
B) 0.2
C) 0.05
D) 0.025
Correct Answer: A
Solution :
\[\underset{2\,mol}{\mathop{2NaHC{{O}_{3}}}}\,\xrightarrow[{}]{\Delta }\underset{1\,mol}{\mathop{N{{a}_{2}}C{{O}_{3}}}}\,+{{H}_{2}}O+C{{O}_{2}}\] \[\because \]2 mol\[NaHC{{O}_{3}}\]on decomposition gives =1 mol \[N{{a}_{2}}C{{O}_{3}}\] \[\therefore \]0.2 mol\[NaHC{{O}_{3}}\]on decomposition will give \[=\frac{1}{2}\times 0.2\] =0.1 mol\[N{{a}_{2}}C{{O}_{3}}\]
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