3Cl2+6 KOH -> KClO3+5KCl+3 H2O. What is the maximum number of moles of KClO3 that could form in the combination of 4.0 moles of Cl2 with 14.0 moles of KOH? no

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When determining the amount of product, the first thing we need to do is to make sure we have a balanced chemical equation.  Since we have that here, we can proceed with the calculations.

3Cl2+6 KOH -> KClO3+5KCl+3 H2O

In order to find out the maximum amount of product, we need to see how many moles of KClO3 could be made based on 4.0 moles of Cl2 and assuming that this is the limiting reagent.

4.0 mol Cl2 (1 mol KClO3 / 3 mol Cl2) = 1.3 mol KClO3

The mole ratio, 1 mol KClO3:3 mol Cl2, comes from the balanced chemical reaction and shows us the relationship between the species in moles. 

Now, we have to make the opposite assumption, that KOH is limiting.

14.0 mol KOH (1 mol KClO3 / 6 mol KOH) = 2.33 mol KClO3

Since assuming that Cl2 was the limiting reagent gave us the smaller number of moles of KClO3, this will be the maximum amount of product formed.  In other words, no matter how much KOH we have, we run out of Cl2 first and cannot make any more KClO3.

 

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