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Molecular Weight of Polymers

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Question

Find out the approximate amount of polythene formed from $20$ kg of calcium carbide from the following reactions.

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

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

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

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Solution

The correct option is B $9$ kg
Molecular weight of $C a C_{2} = 40 + 2 \times 12 = 64$

$∵ 64$ kg of $C a C_{2}$ gives $= 24 + 4 = 28$ kg of ethene.

$∴ 20$ kg of $C a C_{2}$ will give $= \frac{28}{64} \times 20$

$= \frac{7 \times 20}{16} = \frac{35}{4}$

$= 8.75$

$= 9$ kg polythene.

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Similar questions

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}

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

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

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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