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Formation of polyethylene from calcium c...

Formation of polyethylene from calcium carbide takes place as follows
`CaC_(2)+2 H_(2)O rarr Ca(OH)_(2)+C_(2)H_(2)`
`C_(2)H_(2)+H_(2) rarr C_(2)H_(2)`
`N(C_(2)H_(4)) rarr (-CH_(2)-CH_(2)-)_(n)`
The amount of polyethylene obtained from `64.1 kg CaC_(2) ` is

A

7 kg

B

14 kg

C

21 kg

D

28 kg

Text Solution

Verified by Experts

The correct Answer is:
D
`underset(64 g)(CaC_2) + 2H_2O rarr Ca(OH)_2 + C_2H_2`
`C_2H_2+H_2 rarr underset(28g)(C_2H_4)`
64 g of `CaC_2` gives 28g of ethylene
`therefore` 64 kg `CaC_2` will gives 28 kg of polyethylene.
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