A) 11.2
B) 5.6
C) 2.8
D) 8
Correct Answer: A
Solution :
\[\underset{\begin{smallmatrix} 1\,mol \\ 16\,g \end{smallmatrix}}{\mathop{C{{H}_{4}}}}\,+\underset{\begin{smallmatrix} 2\,mol \\ 2\times 22.4L \end{smallmatrix}}{\mathop{2{{O}_{2}}}}\,\xrightarrow{{}}C{{O}_{2}}+2{{H}_{2}}O\] At STP, \[\because \]For the combustion of\[16\text{ }g\text{ }C{{H}_{4}},\]oxygen required\[=2\times 22.4\text{ }L\] \[\therefore \] For the combustion of\[4\text{ }g\text{ }C{{H}_{4}},\]oxygen required will be \[=\frac{2\times 22.4\times 4}{16}L\] \[=11.2L\]
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