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?
PO2 = .953 atmospheres.
The partial pressure of O2(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 O2(g)
|A mass of 3.06 grams of O2(g) has been collected.|
|2.04 grams KClO3 have decomposed.|
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 constant unlike the molar ideal gas R constant does not bother carrying Liters or atmospheres expressly as dimensional factors. Bolztman's k has a dimensional configuration of [Work / Temperature]. R has Liter-atmospheres in its numerator, but we may appreciate the fact that Liter-atmospheres is itself one way of dimensionaly8 exprersssing a quantity of work done.
|.410 moles||.0410 moles|