Formation of polyethylene from calcium carbide takes place as follows CaC 2+2 H 2 O longrightarrow Ca ( OH )2+ C 2 H 2 C 2 H 2+ H 2 longrightarrow C 2 H 4 nC 2 H 4 longrightarrow (- CH 2- CH 2-) n The amount of polyethylene obtained from 64.1 kg of CaC 2 is

Question

Formation of polyethylene from calcium carbide takes place as follows CaC 2+2 H 2 O longrightarrow Ca ( OH )2+ C 2 H 2 C 2 H 2+ H 2 longrightarrow C 2 H 4 nC 2 H 4 longrightarrow (- CH 2- CH 2-) n The amount of polyethylene obtained from 64.1 kg of CaC 2 is (A) 7 Kg (B) 14 Kg (C) 21 Kg (D) 28 Kg

Tardigrade Admin · Accepted Answer

The concerned chemical reactions are
(i)

64 k g C a C_{2} + 2 H_{2} O ⟶ C a (O H)_{2} + Ethyne, 26 k g C_{2} H_{2}

(ii)

C_{2} H_{2} + H_{2} ⟶ Ethylene, 28 k g C_{2} H_{4}

(iii)

n \times 28 k g or 28 k g n C_{2} H_{2} ⟶ n \times 28 k g polythene or 28 k g [- C H_{2} - C H_{2} -] n

Thus

64 k g

of

C a C_{2}

gives

26 k g

of acetylene which in turn gives

28 k g

of ethylene whose

28 k g

gives

28 k g

of the polymer, polythene.

Q. Formation of polyethylene from calcium carbide takes place as follows CaC2​+2H2​O⟶Ca(OH)2​+C2​H2​ C2​H2​+H2​⟶C2​H4​ nC2​H4​⟶(−CH2​−CH2​−)n The amount of polyethylene obtained from 64.1kg of CaC2​ is

7 Kg

14 Kg

21 Kg

28 Kg

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 polyethylene obtained from $64.1 k g$ of $C a C_{2}$ is