Number of Moles

**Jonboy** · May 17th 2008, 05:58 PM

Given: $2KClO_{3}(s) + heat \to 2KCl(s) + 3O_{2}$

How many moles of $O_2$ form as 3.0 mol of $KClO_{3}$ are totally consumed?

I see that the ratio between $O_2$ and $KClO_{3}$ is 3 : 2.

So since $KClO_{3}$ is 3.0 mol, then I figured $O_2$ must be $\boxed{4.5\,moles}$

Is this correct?

**o_O** · May 17th 2008, 07:51 PM

$\text{3.0 mol } KClO_{3} \times \frac{\text{3 mol } O_{2}}{\text{2 mol } KClO_{3} } = \text{4.5 mol } O_{2}$

Looks good to me.

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