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Molecular Weight of Polymers

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Question

Calculate the amount of polythene formed from 20 kg of calcium carbide from the reactions given below :
$C a C_{2} + 2 H_{2} O \to C a (O H)_{2} + C_{2} H_{2}$
$H C \equiv C H + H_{2} P d - B a S O_{4} \to C H_{2} = C H_{2}$
$n C H_{2} = C H_{2} \to (C H_{2} - C H_{2})_{n}$
Polythene

28 g

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6 g

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8.75 kg

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64 kg

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Solution

The correct option is A 8.75 kg
20 kg of $C a C_{2}$ (calcium carbide)
No. of mole of $C a C_{2} = \frac{20 \times 1000}{(40 + 12 + 12)}$
$= \frac{20000}{64} 312.5$ mole
312.5 mol $C a C_{2}$ formed 312.5 mole $C_{2} H_{2}$
312.5 mole $C_{2} H_{2}$ formed $C_{2} H_{4}$ 312.5 mole
Amount of polythene $= 312.5 \times (12 \times 2 + 4)$
$= 312.5 \times 28$
$= 8750 g$
$= 8.75 k g$

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Similar questions

20

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

H C \equiv C H + H_{2} P d \cdot B a S O_{4} - ----- \to C H_{2} = C H_{2}

n C H_{2} = C H_{2} \to (- C H_{2} - C H_{2} -)_{n} P o l y t h e n e

Calculate the amount of polythene formed from 20kg of Calcium Carbide from the reaction given below

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

C H \equiv C H + H_{2} p d . B a S O_{4} ⟶ C H_{2} = C H_{2}

n C H_{2} = C H_{2} ⟶ {(- C H_{2} - C H_{2} -)}_{n} P o l y t h e n e

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 \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}

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