Find out the amount of polythene formed from 20 kg of calcium carbide from the following reactions.
$Ca{C}_2+2H_{2}O⟶Ca{\left(OH\right)}_2+C_2{H}_{2}HC$
$CH\equiv CH+H_2\stackrel{pd.BaS{O}_4}{⟶}C{H}_2=C{H}_2$
$nC{H}_2=C{H}_2⟶\underset{Polythene}{{(-C{H}_2-C{H}_2-)}_n}$

20.0kg of $Ca{C}_2$ $=\frac{20.0\times {10}^3}{64}=312.5$ moles of $Ca{C}_2$
From the above reactions it is evident that 1 mol $Ca{C}_2$ gives 1 mol of acetylene, of which n moles combines with n moles of H2 to give polyethylene (n molecules of $C_2{H}_4$ or polymerized $C_2{H}_4$ ).
Hence 312.5 moles of $Ca{C}_2$ will give 312.5 moles of $C_2{H}_4$ $=312.5\times 28=8750g$ or $8.75kg$
OR the polyethylene formed is $8.75kg$.