# Calculating percentage of use

In the following reaction, $\pu{14g}$ of $\ce{KClO3}$ produces $\pu{1.4g}$ of $\ce{KCl}$ $$\ce{2 KClO3 -> 2 KCl + O2}$$ What is the % of use?

So first we balance the equation: $$\ce{2 KClO3 -> 2 KCl + 3 O2}$$ Next find that $\pu{14g}$ of $\ce{KClO3}$ is $\pu{0.114 mol}$ and $\pu{1.4g}$ of $\ce{KCl}$ is $\pu{0.018 mol}$. We can find how much mols of $\ce{O2}$ we have left (according to conservation of matter) but that doesn't add up with the stoichiometric coefficients, and neither does the previous calculation. So how do I continue from here?

• Note that "% of use" is what is typically called "yield" in chemistry, and "to balance the equation" is preferred over "to even the equation". Regarding the problem, you don't have to take care of oxygen at all, this brings nothing. You know the initial amount and the final amount, that's enough to calculate the yield (%). – andselisk Nov 3 '17 at 10:47
• I thought the formula addresses the total amount of matter. Alright, thanks – Diesel Nov 3 '17 at 11:05
• @TAR86 Please check out the examples in this post. – Jan Nov 3 '17 at 13:40

The question asks you to calculate the yield, i.e. the amount of reactant that was transformed into product. In your example, obviously, not all chlorine atoms that were present in $\ce{KClO3}$ end up in $\ce{KCl}$ since there are $\pu{114mmol}$ of the former but only $\pu{18mmol}$ of the latter although they should be equal (both their stoichiometric coefficients are the same).
$$\text{yield}=\frac{\pu{18mmol}}{\pu{114mmol}}$$