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Standard XII

Chemistry

Other Synthesis Methods

Formation of ...

Question

Formation of polyethene from calcium carbide takes place as follows:
$C a C_{2} + H_{2} O \to C a (O H)_{2} + C_{2} H_{2} C_{2} H_{2} + H_{2} \to C_{2} H_{4}$
$n (C_{2} H_{4}) \to (- C H_{2} - C H_{2} -)_{n}$
The amount of polyethylene possibly obtainable from 64.0 kg $C a C_{2}$ can be:

28 kg

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14 kg

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56 kg

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32 kg

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Solution

The correct option is A 28 kg
64 g $C a C_{2}$ contains 24 g carbon.
All the carbon from calcium carbide is present in polyethylene.
Polyethene is $(C H_{2})_{n}$ , i.e., 14 gram of polythene contains 12 gram of carbon. Hence, 24 gram carbon will be contained in $\frac{14}{12} \times 24 = 28$ g of polyethene.
Therefore, 64 g $C a C_{2}$ will produce 28 g polythene.
Hence, 64 kg $C a C_{2}$ will produce 28 kg polythene.

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Similar questions

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:

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

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C a C_{2}

is :

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

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Q. Formation of polyethylene from calcium carbide takes place as 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}

The amount (in kg) of polyethylene obtained from

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C a C_{2}

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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Formation of polyethene from calcium carbide takes place as follows:CaC2+H2O→Ca(OH)2+C2H2C2H2+H2→C2H4n(C2H4)→(−CH2−CH2−)nThe amount of polyethylene possibly obtainable from 64.0 kg CaC2 can be:

Formation of polyethene from calcium carbide takes place as follows:
$C a C_{2} + H_{2} O \to C a (O H)_{2} + C_{2} H_{2} C_{2} H_{2} + H_{2} \to C_{2} H_{4}$
$n (C_{2} H_{4}) \to (- C H_{2} - C H_{2} -)_{n}$
The amount of polyethylene possibly obtainable from 64.0 kg $C a C_{2}$ can be: