A) 2.5
B) 3
C) 5
D) 7.5
Correct Answer: B
Solution :
\[\underset{=\,36g}{\mathop{\underset{2\,\,\times \,\,18}{\mathop{\underset{2\,mol}{\mathop{2{{H}_{2}}O}}\,}}\,}}\,\xrightarrow{Electrolysis}2{{H}_{2}}\,\,\,\,+\underset{\,\,\,\,\,\,1\,\,mol}{\mathop{\,\,\,\,{{O}_{2}}}}\,\] \[\because \] \[36\,g\]of \[{{H}_{2}}O\] produce 1 mole of oxygen \[\therefore \] 108 g of water will produce oxygen \[=\frac{108}{36}=3\,mol\]
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