A) 0.1
B) 0.2
C) 0.05
D) 0.025
Correct Answer: A
Solution :
\[\underset{2\,\text{mol}}{\mathop{2NaHC{{O}_{3}}}}\,\xrightarrow{\Delta }\underset{1\,\text{mol}}{\mathop{N{{a}_{2}}C{{O}_{3}}}}\,+{{H}_{2}}O+C{{O}_{2}}\] \[\because \] \[\text{2 mol}\,\text{NaHC}{{\text{O}}_{\text{3}}}\]on decomposition gives \[\text{=1}\,\text{mol}\,\text{N}{{\text{a}}_{2}}C{{O}_{3}}\] \[\therefore \] \[\text{0}\text{.2 mol}\,\text{NaHC}{{\text{O}}_{\text{3}}}\] on decomposition will give \[=\frac{1}{2}\times 0.2\] \[\text{=}\,\text{0}\text{.1}\,\text{mol}\,\text{N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\]
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