Home
Class 12
CHEMISTRY
Formation of polyethyene from calcium ca...

Formation of polyethyene from calcium carbide takes place is follows :
`CaC_(2)+2H_(2)OtoCa(OH)_(2)+C_(2)H_(2)`
`C_(2)H_(2)+H_(2)toC_(2)H_(4)`

The amount of polyethylene obtained from `64.0kg " of "CaC_(2)` is

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`
Promotional Banner

Topper's Solved these Questions

  • SYNTHETIC AND NATURAL POLYMERS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises Assertion Reasoning|10 Videos

  • SYNTHETIC AND NATURAL POLYMERS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises Multiple Correct|28 Videos

  • SURFACE CHEMISTRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Subjective|2 Videos

Similar Questions

Explore conceptually related problems

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 form64kg of CaC_(2)

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 64.0 kg CaC_(2) is

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

In C+H_(2)OtoCO+H_(2),H_(2)O acts as

Polythene can be produced from calcium carbide according to the following sequence of reactions. CaC_(2)+2H_(2)Orarr Ca(OH)_(2)+C_(2)H_(2), C_(2)H_(2)+H_(2)rarr C_(2)H_(4), nC_(2)H_(4)rarr(C_(2)H_(4))n The mass of polythene which can be produced from "20.0 kg of "CaC_(2) is :

Balance the following equation: C+H_(2)SO_(4)toCO_(2)+H_(2)O+SO_(2)

Convert C_(2)H_(5)OH to C_(2)H_(5)OCH_(3) ?

CaC_(2)+H_(2)O rarr Ca(OH)_(2)+C_(2)H_(2) 2 mole of CaC_(2) will give y moles of C_(2)H_(2) , % yield is 50% .The value of y is.

CaCl_(2)+Na_(2)C_(2)O_(4) to CaC_(2)O_(4)darr+2NaCl

CaCl_(2)+Na_(2)C_(2)O_(4) to CaC_(2)O_(4)darr+2NaCl