A) \[A{{l}_{4}}{{C}_{3}}\]
B) \[Ca{{C}_{2}}\]
C) VC
D) \[SiC\]
E) \[{{B}_{4}}C\]
Correct Answer: A
Solution :
\[A{{l}_{4}}{{C}_{3}}+12{{H}_{2}}O\to 3C{{H}_{4}}+4Al{{(OH)}_{3}}\] \[Ca{{C}_{2}}+2{{H}_{2}}O\to {{C}_{2}}{{H}_{2}}+Ca{{(OH)}_{2}}\]
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