Polyethylene is obtained from calcium carbide asCaC2+2H2O→Ca(OH)2+C2H2 , C2H2+H2→C2H4 nC2H4→−CH2−CH2nTherefore, the amount of polyethylene obtained from 64 kg CaC2 is ____kg

Moles of CaC2 =64×10364=1×103 ∴From the balanced chemical equation,Moles of C2H4 = moles of C2H2 = moles of CaC2=1×103∴ Moles of polythene =1n×1×103∴ Weight of polythene =1n×1×28nkg=28kg

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Polyethylene is obtained from calcium carbide as
$\begin{array}{l} {CaC}_{2} + 2 H_{2} O \to Ca (OH)_{2} + C_{2} H_{2}, C_{2} H_{2} + H_{2} \to C_{2} H_{4} \\ {nC}_{2} H_{4} \to {(- {CH}_{2} - {CH}_{2})}_{n} \end{array}$
Therefore, the amount of polyethylene obtained from 64 kg CaC₂ is ____kg

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answer is 28.

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Detailed Solution

Moles of CaC₂ $= \frac{64 \times 10^{3}}{64} = 1 \times 10^{3}$
$∴$ From the balanced chemical equation,
Moles of C₂H₄ = moles of C₂H₂ = moles of ${CaC}_{2} = 1 \times 10^{3}$
$∴$ Moles of polythene $= \frac{1}{n} \times 1 \times 10^{3}$
$∴$ Weight of polythene $= \frac{1}{n} \times 1 \times 28 nkg = 28 kg$