| 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
Correct Answer: D
Solution :
| 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. |
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