$Ca{C_2} + 2{H_2}O \to Ca{(OH)_2} + {C_2}{H_2}$
${C_2}{H_2} + {H_2} \to {C_2}{H_4}$
$n({C_2}{H_4}) \to {( - C{H_2} - C{H_2} - )_n}$
$64.1\, kg$ $Ca{C_2}$માંથી મેળવેલ પોલિઇથિલિનનો જથ્થો ......$kg$ છે.
${C_2}{H_2} + {H_2} \to \mathop {{C_2}{H_4}}\limits_{{\rm{28}}\,g} $
$64\,g$ of $Ca{C_2}$ gives $28\,g$ of ethylene
$\therefore $ $64\,kg$ of $Ca{C_2}$ will give $28\,kg$ of polyethylene