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Standard XII

Chemistry

Stoichiometric Calculations

Formation of ...

Question

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 $64.1 k g$ of $C a C_{2}$ is:

7 k g

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14 k g

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21 k g

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28 k g

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Solution

The correct option is D $28 k g$
The molar mass of $C a C_{2}$ is 64 g/mol.

$64 kg$ (or $64000 g$ ) of $C a C_{2} = \frac{64000 g}{64 g/mol} = 1000 mol$ .

1000 moles of $C a C_{2} =$ 1000 moles of $C_{2} H_{2}$
1000 moles of $C_{2} H_{2} =$ 1000 moles of $C_{2} H_{4}$

The molar mass of $C_{2} H_{4}$ is 28 g/mol.

1000 moles of $C_{2} H_{4} = 1000 mol \times 28 g/mol = 28000 g$ or $28 kg$ .

$28 kg$ of $C_{2} H_{4}$ will give $28 kg$ of polythene.

Hence, option D is correct.

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Similar questions

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}

64 kg of

C a C_{2}

will give ______kg of polyethylene.

Q. Polyethylene is obtained from calcium carbide.

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}

Therefore, the amount of polyethylene obtained for 64 kg

C a C_{2}

is:

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:

Q. 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:

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

kg of

C a C_{2}

is :

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Formation of polyethylene from calcium carbide takes place as follows:CaC2+2H2O⟶Ca(OH)2+C2H2C2H2+H2⟶C2H4nC2H4⟶(−CH2−CH2−)nThe amount of polyethene obtained from 64.1 kg of CaC2 is:

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 $64.1 k g$ of $C a C_{2}$ is: