The formation of polyethylene from calcium carbide takes places as follows :
$Ca{C}_2+2H_{2}O\to Ca(OH{)}_2+C_2{H}_2$
$C_2{H}_2+H_2\to C_2{H}_4$
$n{C}_2{H}_4\to (C{H}_2-C{H}_2{)}_n$
The amount of polyethylene obtained from $64$ kg of $Ca{C}_2$ is :

The correct option is B $28$ kg

Mass of $Ca{C}_2$ = $64$ Kg

Moles of $Ca{C}_2$ = 64/64 = $1000$ mol

$Ca{C}_2+2H_{2}O\to Ca(OH{)}_2+C_2{H}_2$

For 1000 mole of $Ca{C}_2$ , We get 1000 mole of $C_2{H}_2$

$C_2{H}_2+H_2\to C_2{H}_4$

For 100 mole of $C_2{H}_2$ , we get 1000 mole of $C_2{H}_4$

$n{C}_2{H}_4\to (C{H}_2-C{H}_2{)}_n$

For n mole $C_2{H}_2$ , we get 1 mole of polyethylene.

For 1000 mole of $C_2{H}_2$ , we get $\left(\frac{1000}{n}\right)\times 1$

Number of moles of polyethylene formed are 1000/n.

Molecular weight of polyethylene is $n\times \left[\right(2\times 12)+(2\times 4\left)\right]=28n$

Mass of polyethylene produced $=$ Number of moles $\times$ Molecular weight

$=\left(\frac{1000}{n}\right)\times 28n$

$=28000$ gram

$=28$ kg

Option D is correct.