Polyethylene is obtained from calcium carbide. CaC_2+2H_2O→CaOH_2+C_2H_2 C_2H_2+H_2→C_2H_4 nC_2H_4→-CH_2-CH_2-n Therefore, the amount of polyethylene…

Question

Polyethylene is obtained from calcium carbide. CaC_2+2H_2O→CaOH_2+C_2H_2 C_2H_2+H_2→C_2H_4 nC_2H_4→-CH_2-CH_2-n Therefore, the amount of polyethylene (in kg) obtained for 64 kg CaC_2 is

MARKS Content Team · Accepted Answer

Correct answer: 28. Moles of CaC_2=(64×10^3)/64≅1×10^3 So from the balanced chemical equation, moles of C_2H_2= moles of C_2H_4 = moles of CaC_2=1×10^3 So moles of polythene =1/n×1×10^3 So weight of polythene =1/n×1×1×28n kg = 28 kg

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 kg $C a C_{2}$ is

Solution

More Polymers Questions

Polyethylene is obtained from calcium carbide. CaC2+2H2O→CaOH2+C2H2 C2H2+H2→C2H4 nC2H4→-CH2-CH2-n Therefore, the amount of polyethylene (in kg) obtained for 64 kg CaC2 is

Solution

More Polymers Questions

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 kg $C a C_{2}$ is