Formation of polyethylene from calcium carbide takes place as follows- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- CaC2 - H2O nbsp- gt- Ca-OH-2 - C2H2 nbsp- gt- C2H4- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- -8203- nbsp- nbsp- -8203-n-C2H4- - gt- - -CH2-CH2- -n-Determine the amount of polyethylene possibly obtainable from 64-0 kg CaCl2 - nbsp-

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Standard XII

Chemistry

Isomerisation of Alkynes

Formation of ...

Question

Formation of polyethylene from calcium carbide takes place as follows:
CaC₂ + H₂O -> Ca(OH)₂ + C₂H₂ -> C₂H₄
n(C₂H₄) -> ( -CH₂-CH₂- )_n
Determine the amount of polyethylene possibly obtainable from 64.0 kg CaCl₂ .

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Solution

Dear Nevadhya,

$I n q u e s t i o n t h e a m o u n t 64 k g o f C a C_{2} i n s t e a d o f C a C l_{2} . T h e s h o r t f o r m o f r e a c t i o n i s : C a C_{2} \to C_{2} H_{4} 64 g C a C_{2} = 28 g C_{2} H_{4} a n d 28 g C_{2} H_{4} w i l l f o r 28 g p o l y t h e n e (n o w t l o s s w h e n e t h e n e f o r m s p o l y e t h e n e) S o a c c o r d i n g l y 64 k g o f C a C_{2} = 28 k g o f C_{2} H_{4} a n d 28 k g o f C_{2} H_{4} w i l l f o r m 28 k g o f p o l y t h e n e .$

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Q. Formation of polyethene from calcium carbide takes place as follows:

C a C_{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}

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Q. Formation of polyethylene from calcium carbide taken place as follows:

C a C_{2} + 2 H_{2} O \to C a (O H)_{2} + C_{2} H_{2}

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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}

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The formation of polyethylene from calcium carbide takes place as follows:

$C a C_{2} + 2 H_{2} O$ $\to$ $C a {(O H)}_{2}$ + $C_{2} H_{2}$
$C_{2} H_{2}$ + $H_{2}$ $\to$ $C_{2} H_{4}$
$n C_{2} H_{4}$ $\to$ ${(- C H_{2} - C H_{2} -)}_{n}$
the amount of polyethylene obtained from 64 kg of $C a C_{2}$ is:

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Formation of polyethylene from calcium carbide takes place as follows: CaC2 + H2O -> Ca(OH)2 + C2H2 -> C2H4 ​ ​n(C2H4) -> ( -CH2-CH2- )n Determine the amount of polyethylene possibly obtainable from 64.0 kg CaCl2 .

Formation of polyethylene from calcium carbide takes place as follows:
CaC₂ + H₂O -> Ca(OH)₂ + C₂H₂ -> C₂H₄
n(C₂H₄) -> ( -CH₂-CH₂- )_n
Determine the amount of polyethylene possibly obtainable from 64.0 kg CaCl₂ .