Chlorine gas was first prepared in 1774 by the oxidation of NaCl with MnO2:

2NaCl (s) + 2H2SO4 (l) + MnO2 (s) →Na2SO4 (s) + MnSO4 (s) + 2H2O (g) + Cl2 (g)

Assume that the gas produced is saturated with water vapor at a partial pressure of 28.7 Hg and that it has a volume of 0.597L at 27 ∘C and 755 mm Hg pressure.

What is the mole fraction of Cl2 in the gas?

0.5

Explanation:

2NaCl (s) + 2H2SO4 (l) + MnO2 (s) → Na2SO4 (s) + MnSO4 (s) + 2H2O (g) + Cl2 (g)

Using ideal gas equation,

PV = nRT

28.7torr

Converting torr to atm,

= 0.0378atm

V = 0.597L

T = 27 °C

= 300 K

a) PV = nRT

(0.0378atm) * (0.597L) = n (0.0821) * (300k)

= 0.000915 mol

moles of water and chlorine = 0.000915 mol

From the above equation, the ratio of water to chlorine = 1 : 2

Therefore, mole of chlorine = 0.000915/2

= 0.000458 mol

mole fraction = moles of specie/moles of all the species present

= 0.000458/0.000915

= 0.5