Small quantities of oxygen gas are sometimes generated by heating KClO3 in the presence of MnO2 HELP?

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 23.0 oC by the reaction of 0.276 g of KClO3 if the barometric pressure is 722.2 mm Hg
The vapor pressure of water at 23.0 oC is 21.1 mm Hg.

Answer 1

Since the oxygen is collected over water, you must subtract out the water vapor pressure to determine the pressure due to just the oxygen.

722.2 mmHg - 21.1 mmHg = 701.1 mmHg = 0.922 atm due to O2

0.276g KClO3 = 0.00175 moles KClO3 (molar mass = 158 g/mol)
Since KClO3 decomposes to form O2 in a 2:3 ratio, the number of moles of O2 produced = 0.00262 moles O2

PV=nRT
(0.922 atm) V = 0.00262 mol (0.08206)(296.2K)

Solving for the volume due to the O2, V = 691 mL

Answer 2

do the stoichometry first. 0.253g KClO3 = ? moles = 0.253 / 122.fifty 5 = 0.002 moles the ratio of KClO3:O2 is two:3 0.002moles KClO3 ( 3 O2/2KClO3) = moles O2 produced = 0.003 moles now, plug this quantity into the suited gasoline regulation p O2 dry = 732.9 - 21.one million = 711.8 mm or 0.936 atm v = ? n = 0.003 r = 0.0821 t = 300K resolve for v = nrt/p v = 0.003 x 0.0821 x 3 hundred / 0.936 = 0.079L or seventy 9 ml