How many milliliters of O2 will form a STP from 55.2 grams of KCLO3?

Answer 1

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 )

Answer 2

Using the molar mass of KClO3, we can find that 55.2 grams is equal to 0.5 moles. Since 2 moles of KClO3 yield 3 moles of O2, we can calculate that 0.5 moles of KClO3 will yield 0.75 moles of O2. Finally, using the ideal gas law at STP (standard temperature and pressure), we find that this is equivalent to 16.8 liters, or 16800 milliliters, of O2.