Question

Polyethylene is obtained from calcium carbide.
$CaC_{2} + 2H_{2}O \to Ca(OH)_{2} + C_{2}H_{2}$
$C_{2}H_{2} +H_{2} \to C_{2}H_{4}$
$ nC_{2}H_{4} \to $
Therefore, the amount of polyethylene obtained for $64 \,kg \,CaC_{2}$ is

• A. $7\, kg$
• B. $14\, kg$
• C. $24\, kg$
• D. $28\, kg$

Answer 1

Answer: (D)

Moles of $CaC_{2} = \frac{64\times 10^{3}}{64} \cong 1\times 10^{3}$
$\therefore $ From the balanced chemical equation,
moles of $C_{2}H_{2}$ = moles of $ C_{2}H_{4}$
= moles of $CaC_{2}= 1 \times 10^{3}$
$\therefore $ moles of polythene $ = \frac{1}{n} \times 1 \times 10^{3}$
$\therefore $ weight of polythene $= \frac{1}{n} \times 1\times28n \, kg = 28\, kg$