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Question

# The formation of polyethylene from calcium carbide takes places as follows :CaC2+2H2O→Ca(OH)2+C2H2C2H2+H2→C2H4nC2H4→(CH2−CH2)nThe amount of polyethylene obtained from 64 kg of CaC2 is :

A
14 kg
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B
7 kg
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C
21 kg
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D
28 kg
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Solution

## The correct option is B 28 kgMass of CaC2 = 64 KgMoles of CaC2 = 64/64 = 1000 molCaC2+2H2O→Ca(OH)2+C2H2For 1000 mole of CaC2 , We get 1000 mole of C2H2 C2H2+H2→C2H4For 100 mole of C2H2 , we get 1000 mole of C2H4 nC2H4→(CH2−CH2)nFor n mole C2H2 , we get 1 mole of polyethylene.For 1000 mole of C2H2 , we get (1000n)×1Number of moles of polyethylene formed are 1000/n.Molecular weight of polyethylene is n×[(2×12)+(2×4)]=28n Mass of polyethylene produced = Number of moles × Molecular weight =(1000n)×28n =28000 gram =28 kgOption D is correct.

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