A) No action takes place
B) \[NaAl{{O}_{2}}\] is formed and \[{{H}_{2}}\] is evolved
C) \[Al{{\left( OH \right)}_{3}}\] is formed and \[{{H}_{2}}\] is evolved
D) \[N{{a}_{2}}Al{{O}_{2}}\] is formed and\[{{H}_{2}}\]is evolved
Correct Answer: B
Solution :
[b] \[Al+NaOH+{{H}_{2}}O\to NaAl{{O}_{2}}+{{H}_{2}}\uparrow .\]
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