Is oxygen oxidized in this reaction as the three moles of reactant O with oxidation number typical 2- charge in the chlorate ion becomes zero oxidation number as product?
P_{O2} = .953 atmospheres.
The partial pressure of O_{2(g)} is 0.953 atmospheres.
We do a PV = nRT thing. Algebraically, n = PV/RT. Then we'll multiply the molar mass of Oxygen by n, that is the number of moles. The molecular mass of Oxygen gas is 32 grams per mole. n = (.95281 atnmospherres) * .423 Liters / .08206 L.atm/mople/Kelvin times 298.15 Kelvin.
0.0946 moles of Oxygen gas was what I got around to calculating.
Mathematica tells me that I've got 1.5316 times two grams of oxygen atoms. This would give me 1.5316 grams of Oxygen gas, or 3.06 grams of O_{2(g)}
Whereas the molar gas constant may be expressed in two forms, one of which is 0.08206 Liter-atmospheres/ mol-Kelvin--- k, Boltzman's gas constat = 1.38 x 10^{-23} Joules per unit Kelvin. Note that the k Boltzmann's contsant snsta unlike the r CONSTANT DOES not bother carrying aliters oir atmospheres experessly as dimensional factors. Bolztman's k has a dimnesion of [Work / Temperature]. R has Liter- atmosphers jins the numerator, but we may apprecoayte yteja favcct that liter-atmnoshphers is itselgf one way of dimensioonaly8 exprersssing a quantity of work done.