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Byju's Answer

Standard XII

Chemistry

Polymer

Formation of ...

Question

Formation of polyethylene from calcium carbide takes place as follows:

$C a C_{2} + 2 H_{2} O ⟶ C a {(O H)}_{2} + C_{2} H_{2}$
$C_{2} H_{2} + H_{2} ⟶ C_{2} H_{4}$
$n C_{2} H_{4} ⟶ {(- C H_{2} - C H_{2} -)}_{n}$

64 kg of $C a C_{2}$ will give ______kg of polyethylene.

7 k g

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14 k g

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21 k g

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28 k g

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Solution

The correct option is D $28 k g$
$C a C_{2} 64.0 k g + 2 H_{2} O ⟶ C a {(O H)}_{2} + C_{2} H_{2} 26 k g$

$C_{2} H_{2} 26 k g + 2 H_{2} ⟶ C_{2} H_{4} 28 k g$

$n C_{2} H_{4} 28 k g ⟶ {(- C H_{2} - C H_{2} -)}_{n} 28 k g$

During addition, nothing is lost.

Therefore, $28 k g$ of $C_{2} H_{4}$ will give $28 k g$ of polyethylene.

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Similar questions

Q. Polyethylene is obtained from calcium carbide.

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

C_{2} H_{2} + H_{2} ⟶ C_{2} H_{4}

n C_{2} H_{4} ⟶ (- C H_{2} - C H_{2} -)_{n}

Therefore, the amount of polyethylene obtained for 64 kg

C a C_{2}

is:

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

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

C_{2} H_{2} + H_{2} ⟶ C_{2} H_{4}

n C_{2} H_{4} ⟶ {(- {C H}_{2} - {C H}_{2} -)}_{n}

The amount of polyethene obtained from

64.1 k g

C a C_{2}

is:

Q. The formation of polyethylene from calcium carbide takes places 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 :

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:

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}

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 64.0 kg of

C a C_{2}

is:

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Formation of polyethylene from calcium carbide takes place as follows:CaC2+2H2O⟶Ca(OH)2+C2H2C2H2+H2⟶C2H4nC2H4⟶(−CH2−CH2−)n64 kg of CaC2 will give ______kg of polyethylene.

The correct option is D 28kgCaC264.0kg+2H2O⟶Ca(OH)2+C2H226kgC2H226kg+2H2⟶C2H428kgnC2H428kg⟶(−CH2−CH2−)n28kgDuring addition, nothing is lost. Therefore, 28kg of C2H4 will give 28kg of polyethylene.

Formation of polyethylene from calcium carbide takes place as follows:

$C a C_{2} + 2 H_{2} O ⟶ C a {(O H)}_{2} + C_{2} H_{2}$
$C_{2} H_{2} + H_{2} ⟶ C_{2} H_{4}$
$n C_{2} H_{4} ⟶ {(- C H_{2} - C H_{2} -)}_{n}$

64 kg of $C a C_{2}$ will give ______kg of polyethylene.