A) \[{{C}_{2}}{{H}_{4}}\]
B) \[{{C}_{3}}{{H}_{4}}\]
C) \[{{C}_{6}}{{H}_{5}}\]
D) \[{{C}_{7}}{{H}_{8}}\]
Correct Answer: D
Solution :
\[{{C}_{x}}{{H}_{y}}+\left( x+\frac{y}{4} \right){{O}_{2}}\to xC{{O}_{2}}+\frac{y}{2}{{H}_{2}}O\] \[x:\frac{y}{2}=\frac{3.08}{44}:\frac{0.72}{18}\] \[\Rightarrow \]\[\frac{2x}{y}=\frac{3.08}{44}\times \frac{18}{0.72}=\frac{7}{4}\]i.e. \[\frac{x}{y}=\frac{7}{8}\]
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