Polyethylene is obtained from calcium carbide. CaC2+2H2O arrow Ca(O H)2+C2H2 C2H2+H2 arrow C2H4 nC2H4 arrow (- C H2 - C H2 -)n Therefore, the amount of polyethylene (in kg) obtained for 64 kg CaC2 is-

Question

Polyethylene is obtained from calcium carbide. CaC2+2H2O arrow Ca(O H)2+C2H2 C2H2+H2 arrow C2H4 nC2H4 arrow (- C H2 - C H2 -)n Therefore, the amount of polyethylene (in kg) obtained for 64 kg CaC2 is- (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Moles of CaC2=(64 × 1 03/64)≅1× 103 So from the balanced chemical equation, moles of C2H2= moles of C2H4 = moles of CaC2=1× 103 So moles of polythene =(1/n)× 1× 103 So weight of polythene =(1/n)× 1× 1× 28 n kg = 28 kg

Q. Polyethylene is obtained from calcium carbide. CaC2​+2H2​O→Ca(OH)2​+C2​H2​ C2​H2​+H2​→C2​H4​ nC2​H4​→(−CH2​−CH2​−)n Therefore, the amount of polyethylene (in kg) obtained for 64kgCaC2​ is-

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

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