5. Potassium chlorate decomposes upon heating as
follows:
2KClO3
(s) → 2KCl(s) + 3O2
(g)
A 2.72-g sample of KClO3 decomposes, and the oxygen at 23.4°C and
0.935 atm is
collected. What volume of oxygen gas will be collected, assuming
100% yield?
2 KClO3 (s) ..............> 2 KCl(s) + 3 O2 (g)
2 mole KClO3 decompose to form 2 mole KCl and 3 mole O2.
2.72-g sample of KClO3 = mass / molar mass = 2.72 g / 122.55 g/mole = 0.0222 mole.
thus
mole of O2 produced = (3 / 2) * 0.0222 = 0.0333 mole.
we have
T = (273 + 23.4) = 296.4 K
P = 0.935 atm
n = 0.0333 mole.
Ideal gas equation is
PV = nRT
or
0.935 * V = 0.0333 * 0.0821 * 296.4
or
V = 0.867 L
volume of oxygen gas will be collected = 0.867 L
5. Potassium chlorate decomposes upon heating as follows: 2KClO3 (s) → 2KCl(s) + 3O2 (g) A...
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