Mole fraction concept

In a certain titration experiment,

1) when $$\ce{KIO3}$$ solution is added to $$\ce{KI}$$, the $$\ce{I-}$$ ion is oxidised to the triiodide complex, $$\ce{I3-}$$

$$\ce{IO3- + 8 I- -> 3 I3- + 3H2O}\tag{1}$$

2) Ascorbic acid immediately reduces the triiodide complex to the colourless $$\ce{I}^-$$ ion.

$$\ce{C6H8O6 + I3^- + H_2O -> C6H8O7 + 3I^- + 2H+}\tag{2}$$

3) When all the ascorbic acid is reacted with $$\ce{I3}^-$$ , the excess $$\ce{I3}^-$$ will react with the starch solution to form a deep blue $$\ce{I3}^-$$ starch complex.

$$\ce{I3^- \text{ + Starch} -> I3^-}\text{- starch complex}\tag{3}$$

At the experiment I have the moles of $$\ce{KIO3}$$ let’s say $$0.1$$ moles.

If I want to find the moles of $$\ce{C6H8O6}$$ can I say that the mole fraction is -> $$\ce{IO3^-} : \ce{I3-} : \ce{C6H8O7}$$ is $$1:3:1$$ ? Meaning the moles of $$\ce{C6H8O7}$$ is $$0.1$$ moles

• Try using equivalents concept. It is more effective – thewitness Nov 7 '18 at 5:41

$$\ce{IO3- + 8 I- -> 3 I3- + 3H2O}\tag{1}$$
$$\ce{3C6H8O6 + 3I3^- + 3H_2O -> 3C6H8O7 + 9I^- + 6H+}\tag{2a}$$
$$\ce{IO3- + 3C6H8O6 -> 3C6H8O7 + I^- + 6H+}\tag{4}$$
So $$\ce{IO3- \text{ to} C6H8O6}$$ is 1::3.