Question

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

Answer 1

Correct Answer - d
`underset(64g)(CaC_(2)+2H_(2))Ooverset(CO or Ni)rarrCa(OH)_(2)+C_(2)H_(2)`
`C_(2)H_(2)+H_(2)rarrunderset(28g)(C_(2)H_(4))`
`64 g` of `CaC_(2)` gives `28 g ` of ethylene
`:. 64 kg` of `CaC_(2)` will give `28 kg ` of polyethylene