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How many milliliters of O2 will form a STP from 55.2 grams of KCLO3?
Wiki User
∙ 13y ago
I assume you mean the decomposition reaction used to produce O2 in lab.
2KCLO3 -> 2KCl + 3O2
find moles O2
55.2 grams KClO3 (1 mole KCLO3/122.55 grams)(3 mole O2/2 mole KClO3)
= 0.67564 moles O2
Now, I use the ideal gas law
PV = nRT
(1 atm)(V) = (0.67564 mol)(0.08206 L*atm/mol*K)(298.15 K)
Volume O2 = 16.53 Liters
which is
16530 milliliters ( less significant figures )
Wiki User
∙ 13y ago
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