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Polymer

Formation of ...

Question

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:

27 kg

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24 kg

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22 kg

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28 kg

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Solution

The correct option is D 28 kg
$M_{C a C_{2}} = 40 + 2 \times 12 = 64 g / m o l$ ,

No. of moles of $C a C_{2}$ in 64 Kg =1000 mole= No. of moles of $C_{2} H_{2}$ formed= No. of moles of $C_{2} H_{4}$ formed

Thus n=1000

Amount of polyethylene $= 28 \times 1000 = 28000 g = 28 k g$

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Similar questions

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 \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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C a C_{2}

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 ⟶ 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}

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}

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