Hydrocarbons Question 204
Question: Formation of polyethylene from calcium carbide takes place as follows $ CaC_2+2H_2O\to Ca(OH)_2+C_2H_2 $ $ C_2H_2+H_2\to C_2H_4 $ $ n(C_2H_4)\to (-CH_2-CH_2-)_n $ The amount of polyethylene obtained from 64.1 kg $ CaC_2 $ is [AIIMS 1997]
Options:
A) 7 kg
B) 14 kg
C) 21 kg
D) 28 kg
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{64g}{\mathop{CaC_2}}+2H_2O\to Ca{{(OH)}_2}+C_2H_2 $
$ C_2H_2+H_2\to \underset{28g}{\mathop{C_2H_4}} $ 64g of $ CaC_2 $ gives 28g of ethylene
$ \therefore $ 64kg of $ CaC_2 $ will give 28kg of polyethylene
BETA
