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Preparation of Hydrogen

The formation...

Question

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:

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 D
28 kg

Polyethylene is formed by this reaction:
$n C_{2} H_{2} + n H_{2}$ $\to$ $(- C_{2} H_{4} -)_{n}$
26 g ethyne forms 28 g of one unit of $(- C_{2} H_{4} -)$
$C a C_{2}$ + $2 H_{2} O$ $\to$ $C_{2} H_{2}$ + $C a {(O H)}_{2}$
64 g calcium carbide forms 26 g ethyne, 64 kg would form 26 kg.
$∴$ 64 kg of $C a C_{2}$ gives 28 kg of ethylene or polyethylene

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Similar questions

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}

C a C_{2}

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}

64

C a C_{2}

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}

64.1 k g

C a C_{2}

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

C a C_{2}

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}

C a C_{2}

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