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)

$28\, kg$

Answer 2

Moles of $CaC_{2} = \frac{64\times 10^{3}}{64} ?$\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$