I have a stoichiometry problem to solve...
I need to find a b and c and x

a KCLOx

HTML Code:

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reacts and form

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--- bKCL + cO2

x could be 1, 2, 3, or 4.
O2 is oxygen gas
Mass of the sample (KCLOx) = 1.0318g
mass of Kcl produced = 0.6251g
Mass of oxygen produced = 0.4067g

I found that
moles of KCL produced is = 0.008385g
moles of O2 produced is = 0.01459g
but I can't find the ratio of moles O2/moles KCL

Thanks

First, the symbol for chlorine is Cl, not CL.

Moles are measured in moles, not in grams.

I get: moles of KCl = 0.6251/(39.1+35.5) = 8.379 x 10^-3mol
moles of O2 = 0.4067/32 = 0.01271 mol
moles of KClOx = 1.0318/(39.1+35.5+ (x*16)) mol

Mole ratio of KCl to O2 = 8.379/0.01271 = 0.659 = 2:3

This gives:

a KClOx -> 2 KCl + 3 O2

a must be 2

2 KClOx -> 2 KCl + 3 O2

So, O must be 3.

ClO3 is the chlorate (V) ion, which exists.