Formation of polyethylene from calcium carbide takes place as follows. CaC2+2H2O→Ca(OH)2+C2H2 C2H2+H2→C2H4 nC2H4→−(CH2−CH2)n− How much of polyethene will be obtained from 16 g of CaC2?

Adding first two equations, we have, CaC2+2H2O+H2→Ca(OH)2+C2H464g 28g16g 7gFrom 3rd equation, nC2H4→−(−CH2−CH2−)n−7 n g of C2H4 gives 7n g polythenei.e., 7g of C2H4 give 7g of polythene.

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Formation of polyethylene from calcium carbide takes place as follows.
$C a C_{2} + 2 H_{2} O \to C a {(O H)}_{2} + C_{2} H_{2} C_{2} H_{2} + H_{2} \to C_{2} H_{4} n C_{2} H_{4} \to - {(C H_{2} - C H_{2})}_{n} -$
How much of polyethene will be obtained from 16 g of $C a C_{2} ?$

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answer is 7.

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Detailed Solution

Adding first two equations, we have,
$\begin{array}{l} C a C_{2} + 2 H_{2} O + H_{2} \to C a {(O H)}_{2} + C_{2} H_{4} \\ 64 g 28 g \\ 16 g 7 g \end{array}$
From 3rd equation,
$n C_{2} H_{4} \to - {(- C H_{2} - C H_{2} -)}_{n} -$
$7 n g$ of $C_{2} H_{4}$ gives $7 n g$ polythene
i.e., $7 g$ of $C_{2} H_{4}$ give $7 g$ of polythene.