Given the following chemical reaction:

6ClO2 (g) + 3H2O (l) > 5HClO3 (l) + HCI (g)

You begin this reaction with 10 moles of ClO2 and 7 moles of H2O. If the amount of water were doubled from 7 moles to 14 moles, how much more HClO3 would be produced?

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6ClO2 (g) + 3H2O(l) --> 5HClO3 (l) + HCl (g)

The question you are asking deals with the concept of the limiting reagent.  We use the balanced chemical equation to compare the moles of each of the reagents to each other to determine which one is available in limited quantity.  We see from the above equation that for every 6 moles of ClO2 consumed, 3 moles of water are required as well.  So the molar ratio of ClO2:H2O is 6:3, which reduces to 2:1.  Now we know that we are starting with 10 moles of ClO2 and 7 moles of H2O.  Let's see which one is the limiting reagent.

10 moles ClO2 * (3 mole H2O/6 mole ClO2) = 5 moles H2O

So in order to react all 10 moles of ClO2, we need 5 moles of H2O to do so.  We have 7 moles of H2O, so that is more than enough.  So that makes ClO2 the limiting reagent.  Let's verify this from the other end: we know that we have 7 moles of H2O.

7 moles H2O * (6 moles ClO2/3 moles H2O) = 14 moles ClO2

So to fully react all 7 moles of water, we would need 14 moles of ClO2.  We only have 10 moles, so that confirms that the limiting reagent is ClO2.  Since ClO2 is the limiting reagent, that means that water is in excess from the very beginning.  So doubling the amount of water from 7 to 14 moles will have no affect on the amount of HClO3 formed since it is only increasing the amount of excess.  10 moles of ClO2 will produce 8.3 moles of HClO3 (multiply 10 by 5/6).  Doubling the amount of water from 7 to 14 moles will not change the amount of HClO3 formed (it will still be 8.3 moles).

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