A) 1 mol\[{{H}_{2}}\]
B) 2 mol\[{{H}_{2}}\]
C) \[\frac{1}{2}\] mol\[{{H}_{2}}\]
D) None of these
Correct Answer: C
Solution :
\[2C{{H}_{3}}OH+\underset{46\,g}{\mathop{2Na}}\,\to \underset{\begin{smallmatrix} sodium \\ ethoxide \end{smallmatrix}}{\mathop{2C{{H}_{3}}ONa}}\,+\underset{1\,mol}{\mathop{{{N}_{2}}}}\,\] \[\because \]\[{{H}_{2}}\]obtains from 46 g Na = 1 mol \[\therefore \]\[{{H}_{2}}\]will obtain from\[23\text{ }g\text{ }Na=\frac{1}{2}mol\]
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