Basic Chem, need 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 20.0 oC by the reaction of 0.261 g of KClO3 if the barometric pressure is 722.7 mm Hg
The vapor pressure of water at 20.0 oC is 17.5 mm Hg.

How do you do this?

Answer 1

First of all let's find how many moles of KClO3 you have and how many moles of O2 are produced

mole KClO3=mass/MW =0.261/122.5 =2.13*10^-3

2KClO3(s) → 2KCl(s) + 3O2(g)
thus 2mole KClO3 produce 3 mole O2
2.13*10^-3 mole KClO3 produce x mole O2

=> 2x=3*2.13*10^-3 => x=3.19*10^-3 mole O2

In order to find the volume you use PV=nRT, but how much is P?
You know that the total pressure is 722.7 mm Hg and the pressure comming from water is 17.5 mm Hg. Thus the pressure coming from O2 is 722.7-17.5= 705.2 mm Hg.
You have to convert into atm.
1 atm is 760.0 mm Hg
Thus 705.2 mm Hg is 705.2/760.0 = 0.928 atm. This value of P will go in the equation

PV=nRT=> V=nRT/P= (3.19*10^-3)*0.082*293/0.928= 0.08259 L= 82.59ml