• # 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

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