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  4. Polyethylene is obtained from calcium carbide. CaC2+2H2O arrow Ca(O H)2+C2H2 C2H2+H2 arrow C2H4 nC2H4 arrow (- C H2 - C H2 -)n Therefore, the amount of polyethylene (in kg) obtained for 64 kg CaC2 is-

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Q. Polyethylene is obtained from calcium carbide.
$CaC_{2}+2H_{2}O \rightarrow Ca\left(O H\right)_{2}+C_{2}H_{2}$
$C_{2}H_{2}+H_{2} \rightarrow C_{2}H_{4}$
$nC_{2}H_{4} \rightarrow \left(- C H_{2} - C H_{2} -\right)n$
Therefore, the amount of polyethylene (in kg) obtained for $64\, kg \,CaC_{2}$ is-

NTA AbhyasNTA Abhyas 2022

Solution:

Moles of $CaC_{2}=\frac{64 \times 1 0^{3}}{64}\cong1\times 10^{3}$
So from the balanced chemical equation, moles of $C_{2}H_{2}=$ moles of $C_{2}H_{4}$
$=$ moles of $CaC_{2}=1\times 10^{3}$
So moles of polythene $=\frac{1}{n}\times 1\times 10^{3}$
So weight of polythene $=\frac{1}{n}\times 1\times 1\times 28\,n$ kg
$= 28 \,kg$