A) \[\text{3 }mL\]
B) \[12\text{ }mL\]
C) \[\text{24 }mL\]
D) \[\text{48 }mL\]
Correct Answer: D
Solution :
\[\underset{12\,mL}{\mathop{\underset{1\,mol}{\mathop{{{C}_{4}}{{H}_{10}}}}\,}}\,+\frac{13}{2}{{O}_{2}}\xrightarrow{{}}\underset{12\times 4=48\,\,mL}{\mathop{\underset{4\,mol}{\mathop{4C{{O}_{2}}}}\,}}\,+5{{H}_{2}}O\]
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