A) \[{{C}_{2}}{{H}_{6}}\]
B) \[{{C}_{2}}{{H}_{4}}\]
C) \[{{C}_{3}}{{H}_{6}}\]
D) \[{{C}_{3}}{{H}_{8}}\]
Correct Answer: D
Solution :
[d] Let the formula of hydrocarbon A is CxHy. \[{{C}_{x}}{{H}_{y}}(g)+\left( x+\frac{y}{4} \right){{O}_{2}}(g)\xrightarrow{{}}xC{{O}_{2}}(g)+\frac{y}{2}{{H}_{2}}O(l)\]\[1\,mL\,\,\,\,\,\,\,\,\,\,\,\left( x+\frac{y}{4} \right)mL\] \[xmL\,\,\,-\] \[15\,mL\,\,\,\,\,\,\,\,\,\,\,15\left( x+\frac{y}{4} \right)mL\] \[15x\,mL-\] Volume of \[{{O}_{2}}\frac{357\times 21}{100}=75\,mL;\] volume of \[{{N}_{2}}\] \[=357-75=282\text{ }mL\] Volume of \[C{{O}_{2}}=327-282=45mL\] \[\therefore \,\,\,\,\,\,\,\,15x=45,\,\,x=3\] \[\therefore \,\,\,\,\,\,\,\,\left( \begin{align} & 15\left( x+\frac{y}{4} \right)=75 \\ & x+\frac{y}{4}=5 \\ \end{align} \right)\] \[3+\frac{y}{4}=5,\]On solving we get\[y=8\]. Formula of \[(A)={{C}_{3}}{{H}_{8}}\]
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