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The fomation of polyethylene from calciu...

The fomation of polyethylene from calcium carbide takes place as follows:
`CaC_(2)+2H_(2)OrarrCa(OH_(2))+C_(2)H_(2)`
`C_(2)H_(2)+H_(2)rarrC_(2)H_(4)`

The amount of polyethylene obtained form`64kg of`CaC_(2)`

A

`14kg`

B

`7kg`

C

`21kg`

D

`28kg`

Text Solution

Verified by Experts

The correct Answer is:
D
n mol of `CaC_(2)`=n mol of`C_(2)H_(4)`=n mol of
`nxx64kg=nxx28kg=nxx28kg`
Amount of polyethene=`28kg`
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