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**How many grams of potassium chlorate (KClO3) decompose to potassium chloride and 638 mL of O2 at 128 C and 752 torr?**

2KClO_{3}(s) 2KCl (s) + 3O_{2}(g)I use P = nRT/v, therefore n = PV/RT

R=.0821

V= .638 L

T = 128+375.15 = 503.15 K

P = 752 torr x (1atm/760torr) = .989 atm

(.989 atm x .638 L)/(.0821 x 503.15 K) = .0152 mol (n)

So .0152 mol x (122.55 g KClO

_{3}/1mol) = 1.8627 g

OR is not?