13, Formation of polyethylene, from calcium carbide,takes place as follows:CaC2 + 2H2OCa(OH)2 + C2H2C2H2 + H2n(C2H4)(-CH2-CH2-)n.C2H4 ;The amount of polyethylene obtained from 64 kg ofCaC is(1) 7 kg(3) 21 kg(2) 14 kg(4),28 kg

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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. 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 :

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. 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

64.1

kg of

C a C_{2}

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.

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