JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer At 300 K and 1 atmospheric pressure, 10 Ml of a hydrocarbon required 55 mL of \[{{O}_{2}}\] for complete combustion and 40 mL of \[C{{O}_{2}}\] is formed. The formula of the hydrocarbon is : [JEE Main 10-4-2019 Morning]

    A) \[{{C}_{4}}{{H}_{8}}\]                 

    B) \[{{C}_{4}}{{H}_{7}}Cl\]

    C) \[{{C}_{4}}{{H}_{10}}\]               

    D) \[{{C}_{4}}{{H}_{6}}\]

    Correct Answer: D

    Solution :

    \[{{C}_{x}}{{H}_{y}}+\left( x+\frac{y}{4} \right){{O}_{2}}\xrightarrow[{}]{{}}xC{{O}_{2}}+\frac{y}{2}{{H}_{2}}O\]           \[10\,\,\,\,\,\,\,\,10\left( x+\frac{y}{4} \right)\,\,\,\,\,\,\,\,10x\] By given data,\[10\left( x+\frac{y}{4} \right)=55\]                ?.(1) \[10x=40\]                                           .... (2) \[\therefore x=4,y=6\Rightarrow {{C}_{4}}{{H}_{6}}\]

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