Question

Formation of polyethylene from calcium carbide takes place as follows
$ CaC _{2}+2 H _{2} O \longrightarrow Ca ( OH )_{2}+ C _{2} H _{2}$
$C _{2} H _{2}+ H _{2} \longrightarrow C _{2} H _{4} $
$ nC _{2} H _{4} \longrightarrow \left(- CH _{2}- CH _{2}-\right) n$
The amount of polyethylene obtained from $64.1\, kg$ of $CaC _{2}$ is

• A. 7 Kg
• B. 14 Kg
• C. 21 Kg
• D. 28 Kg

Answer 1

Answer: (D)

The concerned chemical reactions are
(i) $\underset{64\,kg}{CaC _{2}}+2 H _{2} O \longrightarrow Ca ( OH )_{2}+ \underset{\text{Ethyne}, 26\,kg}{C _{2} H _{2}}$
(ii) $C _{2} H _{2}+ H _{2} \longrightarrow \underset{\text{Ethylene},28\,kg}{C _{2} H _{4}}$
(iii) $\underset{n\times 28\,kg \,or\, 28\,kg}{nC _{2} H _{2}} \longrightarrow \underset{n\times 28\,kg\, \text{polythene or }28\,kg}{\left[- CH _{2}- CH _{2}-\right] n}$
Thus $64 \,kg$ of $CaC _{2}$ gives $26\, kg$ of acetylene which in turn gives $28\, kg$ of ethylene whose $28\, kg$ gives $28\, kg$ of the polymer, polythene.