Question

At elevated temperatures, solid sodium chlorate NaClO3 decomposes to produce sodium chloride, NaCl, and O2 gas. A 0.8948 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 63.2 mL at 23.0 C and 738 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.

Answer 1

Vapor pressure of water= 21.07 Torr

Partial pressure of oxygen, P = 738-21.07=716.93 Torr =716.93/760 atm=0.9433atm

Volume, V = 63.2ml =63.2/1000L=0.0632L, T=23 deg.c= 23+273.15=296.15K, R= 0.08206 L.atm/mole.K

moles of oxygen, n = PV/RT= 0.9433*0.0632/ (0.08206*296.15)=0.002453 moles, mass of oxygen = 0.78501 gms

The decomposition of NaClO3 can be represented as 2NaClO3--->2 NaCl+3O2

3 moles of oxygen correspond to 2 moles of NaClO3

0.002453 moels of oxygen correspond to 0.002453*2/3=0.001635 moles of NaClO3

Molecular weight of NaClO3= 23+35.5+48=106.5 g/mol. mass of NaClO3= 0.001635*106.5=0.174gm

mass percentage of NaClO3 in the sample = 100*0.174/0.8948=19.445%