Formation of polyethylene from calcium carbide takes place as follows
`CaC_(2)+2 H_(2)O rarr Ca(OH)_(2)+C_(2)H_(2)`
`C_(2)H_(2)+H_(2) rarr C_(2)H_(2)`
`N(C_(2)H_(4)) rarr (-CH_(2)-CH_(2)-)_(n)`
The amount of polyethylene obtained from `64.1 kg CaC_(2) ` is

Options

A. 7 kg
B. 14 kg
C. 21 kg
D. 28 kg

Answer

D

Explanation

The concerned chemical reactions are
(i) `underset(64kg)(CaC_2) + 2H_2O to Ca(OH)_2 + underset("Ethyne, 26kg")(C_2H_2)`
(ii) `C_2H_2 + H_2tounderset("Ethylene, 28kg")(C_2H_4)`
(iii) `underset("or 28kg")underset(nxx28kg)(nC_2H_2) to underset(or 28 kg)underset(n xx28kg " polythene")([-CH_2-CH_2-]_n)`
Thus 64 kg of `CaC_2` gives 26 kg of acetylene which in turn gives 28 kg of ethylene whose 28 kg gives 28 kg of the polymer, polythene.