Question

Formation of polyethylene from calcium carbide takes place as follows

\(CaC_2 +2H_2O \longrightarrow Ca(OH)_2+C_2H_2\)

\(C_2H_2+H_2 \longrightarrow C_2H_4\)

The amount of ethylene obtained from 64.1 kg CaC₂ is

(a) 7 kg

(b) 14 kg

(c) 21 kg

(d) 28 kg

Answer 1

(d) 28 kg

The molar mass of CaC₂​ is 64 g/mol.

64 kg (or 64000 g) of CaC₂ ​= \(\frac{64000\,g}{64\,g/mol}=1000\,mol.\)

1000 moles of CaC₂​ = 1000 moles of C₂​H_2​

1000 moles of C₂​H₂ ​= 1000 moles of C_2​H_4​

The molar mass of C₂​H₄​ is 28 g/mol.

1000 moles of C₂​H₄​ = 1000 mol x 28 g/mol = 28000 g or  28 kg.

28 kg of C₂​H_4​ will give 28 kg of polythene.