A) Butane
B) Isobutane
C) Ethane
D) Propane
Correct Answer: D
Solution :
\[{{C}_{n}}{{H}_{2n+2}}+\left( \frac{3n+1}{2} \right){{O}_{2}}\xrightarrow[{}]{{}}nC{{O}_{2}}+(n+1){{H}_{2}}O\] 5 L 25 L Since volumes are measured at constant T & P So, Volume ? mole \[\therefore \]\[{{n}_{alkane}}=\left( \frac{2}{3n+1} \right)\times {{n}_{{{O}_{2}}}}\] \[5=\frac{2}{3n+1}\times 25\] \[\therefore \]\[n=3\] \[\therefore \]Alkane is propane \[({{C}_{3}}{{H}_{8}}).\].
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