Question

Polyethylene can be produced from $Ca{C}_2$ according to the following sequence of reactions.
$Ca{C}_2+H_{2}O\to CaO+C_2{H}_2\text{(Yield = 70 \%)}$
$C_2{H}_2+H_2\to C_2{H}_4\text{(Yield = 50 \%)}$ $n{C}_2{H}_4\to (C{H}_{2}C{H}_2{)}_n\text{(Yield = 20 \%)}$
Calculate the mass of polyethylene which can be produced from $20\text{kg}$ of $Ca{C}_2$.

• A. $612\text{kg}$
• B. $558\text{g}$
• C. $612\text{g}$
• D. $558\text{kg}$

Answer 1

The correct option is C $612\text{g}$

$Ca{C}_2+H_{2}O\to CaO+C_2{H}_2\text{(i)}$
$C_2{H}_2+H_2\to C_2{H}_4\text{(ii)}$
$n{C}_2{H}_4\to (C{H}_{2}C{H}_2{)}_n\text{(iii)}$
In reaction (i),
$64\text{g}$ of $Ca{C}_2$ produces $26\text{g}$ of $C_2{H}_2$
$20\text{kg}$ will produce $=\frac{26}{64}\times 20\times {10}^3=8.125\times {10}^3\text{g}$
Yield of the reaction is given as 70 %
Actual amount formed $=8.125\times {10}^3\times 0.7=5.687\times {10}^3\text{g}$
In reaction (ii),
$26\text{g}$ of $C_2{H}_2$ produces $28\text{g}$ of $C_2{H}_4$
$5.687\times {10}^3\text{g}$ will produce = $\frac{28}{26}\times 5.687\times {10}^3=6.125\times {10}^3\text{g}$
Yield of the reaction is given as 50 %
Actual amount of $C_2{H}_4$ formed $=6.125\times {10}^3\times 0.50=3.06\times {10}^3\text{g}$
In reaction (iii),
$28\times n\text{g}$ of $C_2{H}_4$ forms $28\times n\text{g}$ of polyethylene
$3.06\times {10}^3\text{g}$ will form $=3.06\times {10}^3\text{g}$ of polyethylene
Yield of the reaction is given as 20 %
Actual yield $=3.06\times {10}^3\times 0.20=612\text{g}$