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JEE Main & Advanced Chemistry Hydrocarbons / हाइड्रोकार्बन Question Bank Critical Thinking

  • question_answer
    Formation of polyethylene from calcium carbide takes place as follows \[Ca{{C}_{2}}+2{{H}_{2}}O\to Ca{{(OH)}_{2}}+{{C}_{2}}{{H}_{2}}\] \[{{C}_{2}}{{H}_{2}}+{{H}_{2}}\to {{C}_{2}}{{H}_{4}}\] \[n({{C}_{2}}{{H}_{4}})\to {{(-C{{H}_{2}}-C{{H}_{2}}-)}_{n}}\] The amount of polyethylene obtained from 64.1 kg \[Ca{{C}_{2}}\] is [AIIMS 1997]

    A) 7 kg

    B) 14 kg

    C) 21 kg

    D) 28 kg

    Correct Answer: D

    Solution :

               \[\underset{\text{64}\,g}{\mathop{Ca{{C}_{2}}}}\,+2{{H}_{2}}O\to Ca{{(OH)}_{2}}+{{C}_{2}}{{H}_{2}}\]            \[{{C}_{2}}{{H}_{2}}+{{H}_{2}}\to \underset{\text{28}\,g}{\mathop{{{C}_{2}}{{H}_{4}}}}\,\]            64g of \[Ca{{C}_{2}}\] gives 28g of ethylene \[\therefore \] 64kg of \[Ca{{C}_{2}}\] will give 28kg of polyethylene


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