Formation of polyethylene from calcium carbide takes place as given below:
$Ca{C}_2+2H_{2}O\to Ca{\left(OH\right)}_2+C_2{H}_2$
$C_2{H}_2+H_2\to C_2{H}_4$
$n{C}_2{H}_4\to {(-C{H}_2-C{H}_2-)}_n$
The amount (in kg) of polyethylene obtained from $64.1$ kg of $Ca{C}_2$ is

The correct option is C $28$
Number of moles of $Ca{C}_2=\frac{64.1\times {10}^3}{64}=1000$ mol

One mole of $Ca{C}_2$ on hydrolysis and hydrogenation forms one mole of ethene.

$\therefore$ Number of moles of ethene formed $=1000$ mol

$\therefore$ Mass of polyethene $=(1000\times 28)$ g $=28$ kg