Chemistry Question on Stoichiometry and Stoichiometric Calculations

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

Answer 1

Solution

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.