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## Homework Statement

2 KClO3 (s) + heat --> 2 KCl (s) + 3 O2 (g)

If 0.20 g of KClO3 (s) decomposes in 2.8 s, at what rate is oxygen O2 (g) released over this time at SATP?

## Homework Equations

n = PV/RT

## The Attempt at a Solution

Converted KClO3 to mol.

.20 x 1 / 122.5495g = 0.001632 mol KCl

Find rate of decomposition :

.001632 mol in 2.8 sec --> 2 mol in 3431.37 sec

2/3431.37 = .000583 mol/s

The rate of decomposition of KClO3 is .000583 mol/s

I'm stuck at this part. I know I need to use some sort of n = PV/RT since it tells me that it is at SATP. Can I have some hints or help?