What volume (ml) of O2 is collected over water at 24.0 oC by the reaction of 0.285 g of KClO3?

Question

Small quantities of oxygen gas are sometimes generated by heating KClO3 in the presence of MnO2 as a catalyst:



2KClO3(s) → 2KCl(s) + 3O2(g)



What volume (ml) of O2 is collected over water at 24.0 oC by the reaction of 0.285 g of KClO3 if the barometric pressure is 729.9 mm Hg
The vapor pressure of water at 24.0 oC is 22.4 mm Hg.

Answer 1

Change grams to moles: 0.285 g/122.5g/mole = 0.00233 moles

Change to moles O2 (ratio from equation is 3/2)
0.00233 x 3/2 = 0.00349 moles O2
Calculate vol. from relationship 22.4L/mole for any gas at STP
22.4L/mole x 00349 mole = 0.0782 L at STP

Use Combined Gas Law to change to vol. at 24.0 degrees and 729.9mm. Reduce pressure by the water vapor pressure.

P1 = 760.mm V1 = 0.0782 T1 = 0 + 273
P2 = 729.8-22.4 V2 = x T2 = 24 + 273

x = (0.0782)(760)(297)/(273)(707.4) = 0.0914 L or 91.4 mL