A) 27 kg
B) 24kg
C) 22 kg
D) 28 kg
Correct Answer: D
Solution :
(d) Adding first two equations, we have, -\[\underset{64g}{\mathop{\text{Ca}{{\text{C}}_{\text{2}}}}}\,+\text{2}{{\text{H}}_{\text{2}}}\text{O}\xrightarrow{{}}\text{Ca(OH}{{\text{)}}_{\text{2}}}+\underset{28g}{\mathop{{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}}}\,\] i.e. 64 g of \[\text{Ca}{{\text{C}}_{\text{2}}}\]gives 28 g of \[{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\] From \[\text{III rd}\]reaction \[\underset{n\times 28ng}{\mathop{n{{C}_{2}}{{H}_{4}}}}\,\xrightarrow{{}}-\underset{28ng}{\mathop{(C{{H}_{2}}}}\,-C{{H}_{2}})\underset{n}{\mathop{-}}\,\] Here, 28 n g of \[{{C}_{2}}{{H}_{4}}\]gives 28n g of polythene, i.e. 28n g of \[{{C}_{2}}{{H}_{4}}\] gives 28 g of polythene. Thus 64 g of \[Ca{{H}_{2}}\] will give 28 g of polyethylene.
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