Question

Formation of polyethene from calcium carbide takes place as follows :

CaC2+H2O $\rightarrow$Ca(OH)2 + C2H2 ; C2H2 + H2$\rightarrow$ C2H4

n(C2H4) $\rightarrow$ (-CH2-CH2-)n.

The amount of polyethylene possibly obtainable from 64.0 kg CaC2 can be

• A. 28Kg
• B. 14kg
• C. 21kg
• D. 42 kg

Answer 1

Answer: (A)

28Kg

Answer 2

(1)

CaC2 + H2O $\longrightarrow$Ca(OH)2 + C2H$\longrightarrow$ C2H4 ... (1)

nC2H4 $\longrightarrow$ ... (2)

From equation (1)

mole of CaC2 = mole of C2H4

$\frac{64 \times 10^{3}}{64}$ = mole of C2H4

From equation (2)

$\frac{moleofC_{2}H_{4}}{n} = \frac{moleofpolymer}{1}$

$\frac{10^{3}}{n} = \frac{wtofpolymer}{n(28)}$

wt of polymer = 28 × 103 g = 28 Kg