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Standard XII

Chemistry

Head Elements (Group 1)

Polyethylene ...

Question

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:

7 kg

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14 kg

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21 kg

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28 kg

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Solution

The correct option is C 28 kg
$C a C_{2} + 2 H_{2} O ⟶ C a {(O H)}_{2} + C_{2} H_{2}$ ,
64 kg $C a C_{2}$ = 1000 mol ( $∵$ 64 is molecular wt. of $C a C_{2}$ )
Here 1000 mol of $C a C_{2}$ gives 1000 mol of acetylene and it gives 1000 mol of ethylene. All of this ethylene is used to make polyethylene.
$S o m o l = \frac{W t .}{M o l e c u l a r W t .}$
$1000 = \frac{W t .}{28}$
$W t . = 28 \times 1000$
$W t . = 28 k g$
Therefore, the amount of polyethylene obtained for 64 kg $C a C_{2}$ is 28kg.

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Similar questions

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}

64 kg of

C a C_{2}

will give ______kg of polyethylene.

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 :

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:

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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Polyethylene is obtained from calcium carbide.CaC2+2H2O⟶Ca(OH)2+C2H2C2H2+H2⟶C2H4nC2H4⟶(−CH2−CH2−)nTherefore, the amount of polyethylene obtained for 64 kg CaC2 is:

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: