Ok Ok...

Question:

Find the number of moles [AFTER Reaction] of KBrO3

Given:

initial moles of KBrO3 [before titration reaction] = 0.0004

BrO3(-) + 9I(-) + 6H30(+) ----%26gt; Br(-) + 3I3(-) +9H20

I3(-) + 2S2O3(2-) ---%26gt; 3I(-) + S4O6(2-)

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NOTE::: The above equations ARE ALREADY BALANCED

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My teacher said that,

"To find the number of moles of bromate reacted, subtract the number of moles of bromate remaining from the initial number of moles of bromate. Find the number of moles of bromate remaining from the moles of thiosulfate used in the titration and the relationship between the moles of bromate and the moles of thiosulfate as shown by the two equations "

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So i have no idea how to do that but can some please show me how?

Thank you for your consideration.

How do you get Moles ratio?

This is extremely simple. You do not give any numbers but this is how you will do it. (You will probably be given data on the quantitative results of the second reaction and find out how much reacted). But from your question you need to do this:

Let say you have 6 moles of S2O3(=). From the equation you should note that 1 mole I3(-) require 2 moles of S2O3(=). So if you have 6 moles of S2O3(=) you will have 3 moles of I3(-). From the first reaction 3 moles of I3(-) will require 1 mol of BrO3(-).

So 1 mol BrO3(-) will "require" 6 moles S2O3(=).

The number you are not giving is how many moles of S2O3(=) you were given in the question or need to calculate from data given about the titration.

Assume that it is x moles of S2O3(=).

Then the number of moles of BrO3(-) that reacted will be x/6.

So the answer will be 0.0004 - 6/x

dendrobium

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