• question_answer Formation of polyethylene from calcium carbide takes place as follows $Ca{{C}_{2}}+2{{H}_{2}}O\xrightarrow{{}}Ca{{(OH)}_{2}}+{{C}_{2}}{{H}_{2}}$ ${{C}_{2}}{{H}_{2}}+{{H}_{2}}\xrightarrow{{}}{{C}_{2}}{{H}_{4}}$ $n{{C}_{2}}{{H}_{4}}\xrightarrow{{}}{{\left( -C{{H}_{2}}-C{{H}_{2}}- \right)}_{n}}$ The amount of polyethylene obtained from 64.1 kg of $Ca{{C}_{2}}$is A) 7 kg B) 14 kg C) 21 kg D) 28 kg

 The concerned chemical reactions are (i) $\underset{64kg}{\mathop{Ca{{C}_{2}}}}\,+2{{H}_{2}}O\to Ca{{\left( OH \right)}_{2}}+\underset{Ethyne,26kg}{\mathop{{{C}_{2}}{{H}_{2}}}}\,$ (ii) ${{C}_{2}}{{H}_{2}}+{{H}_{2}}\to \underset{Ethylene,28kg}{\mathop{{{C}_{2}}{{H}_{4}}}}\,$ (iii) $\underset{\begin{smallmatrix} n\times 28kg \\ or\,28kg \end{smallmatrix}}{\mathop{n{{C}_{2}}{{H}_{4}}}}\,\to \underset{\begin{smallmatrix} n\times 28kg\,polythene \\ or\,28kg \end{smallmatrix}}{\mathop{{{\left[ -C{{H}_{2}}-C{{H}_{2}}- \right]}_{n}}}}\,$ Thus 64 kg of ${{\operatorname{CaC}}_{2}}$ gives 26 kg of acetylene which in turn gives 28 kg of ethylene whose 28 kg gives 28 kg of the polymer, polythene.