Polyethylene is obtained from calcium carbide. CaC2 + 2H2O → Ca(OH)2 + C2H2 C2H2 +H2 → C2H4 nC2H4 → Therefore, the amount of polyethylene obtained for 64 kg CaC2 is

Question

Polyethylene is obtained from calcium carbide. CaC2 + 2H2O → Ca(OH)2 + C2H2 C2H2 +H2 → C2H4 nC2H4 → <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/9621c3152d386bb536b9b67d6d4686c9-.png" /> Therefore, the amount of polyethylene obtained for 64 kg CaC2 is (A) 7 kg (B) 14 kg (C) 24 kg (D) 28 kg

Tardigrade Admin · Accepted Answer

Moles of CaC2 = (64× 103/64) ≅ 1× 103 ∴ From the balanced chemical equation, moles of C2H2 = moles of C2H4 = moles of CaC2= 1 × 103 ∴ moles of polythene = (1/n) × 1 × 103 ∴ weight of polythene = (1/n) × 1×28n kg = 28 kg

Q. Polyethylene is obtained from calcium carbide.
$C a C_{2} + 2 H_{2} O \to C a (O H)_{2} + C_{2} H_{2}$
$C_{2} H_{2} + H_{2} \to C_{2} H_{4}$
$n C_{2} H_{4} \to$
Therefore, the amount of polyethylene obtained for $64 k g C a C_{2}$ is

$7 k g$

$14 k g$

$24 k g$

$28 k g$

Q. Polyethylene is obtained from calcium carbide. CaC2​+2H2​O→Ca(OH)2​+C2​H2​ C2​H2​+H2​→C2​H4​ nC2​H4​→ Therefore, the amount of polyethylene obtained for 64kgCaC2​ is

7kg

14kg

24kg

28kg

Moles of CaC2​=6464×103​≅1×103 ∴ From the balanced chemical equation, moles of C2​H2​ = moles of C2​H4​ = moles of CaC2​=1×103 ∴ moles of polythene =n1​×1×103 ∴ weight of polythene =n1​×1×28nkg=28kg

Q. Polyethylene is obtained from calcium carbide.
$C a C_{2} + 2 H_{2} O \to C a (O H)_{2} + C_{2} H_{2}$
$C_{2} H_{2} + H_{2} \to C_{2} H_{4}$
$n C_{2} H_{4} \to$
Therefore, the amount of polyethylene obtained for $64 k g C a C_{2}$ is

$7 k g$

$14 k g$

$24 k g$

$28 k g$