Question

At elevated temperatures, solid sodium chlorate NaClO3decomposes to produce sodium chloride, NaCl, and O2 gas. A 0.9560 g sample of impure sodium chlorate was heated until the production of oxygen ceased. The oxygen gas was collected over water and occupied a volume of 63.6 mL at 23.0 ∘C and 737 Torr .

Calculate the mass percentage of NaClO3 in the original sample. Assume that none of the impurities produce oxygen on heating. The vapor pressure of water is 21.07 Torr at 23 ∘C.

Answer 1

Decomposition of NaClO3

2NaClO₃(s) ---> 2NaCl(s)+ 3O₂(g)

From Dalton's law, pressure due to O₂

P_tot = P_O2 + P_H2O                

737 torr=P_O2+21.07 torr           

      P_O2= 715.93 torr(1 atm/760 torr)= 0.942 atm

Volume of gas

      V= 63.6mL= 0.0636 L

Temperature

T =23^oC + 273= 300 K

from the ideal gas equation

Moles of gas

PV=nRT

(0.942atm)(0.0636L)=n(0.0821 atm L mol^-1K^-1)(300K)

n=0.002432 mol

From the stoichiometry of the reaction

Moles of NaClO3 consumed

= 0.002432 mol O2 x 2 mol NaClO3 / 3 mol O2

= 0.00162 mol NaClO3

Mass of NaClO3 = moles x molecular weight

= 0.00162 mol x 106.5g/mol

= 0.1727 g

% mass of NaClO3 = (0.1727g/0.9560g) x 100%

=18.065%