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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Zoloft on January 23, 2009, 06:44:45 AM

Title: How does my solution look...
Post by: Zoloft on January 23, 2009, 06:44:45 AM

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

2KClO₃(s) :rarrow: 2KCl (s) + 3O₂(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 KClO₃/1mol) = 1.8627 g

OR is not?

Title: Re: How does my solution look...
Post by: sjb on January 23, 2009, 07:20:11 AM

Looks fine, but check the temperature :)

S