Why moles of (S2O3 2-) is equal to mol of ( MnO4-) times 5

[mimi88]

Homework Statement

The ratio of KMnO4 to I2 is given by
2 MnO4 - + 16 H+ + 10 I - ⇌ 2 Mn2+ + 5 I2 + 8 H2O
the ratio is 2:5

- The ratio of I2 to S2O3
2- is given by 1 I2 + 2 S2O32- ⇌ 2 I-+ S4O5 2-
the ratio is 1:2

therefore n(S2O3 2- ) = n(MnO4 - ) x 5

Homework Equations

Why the mole of (S2O3 2-) is equal to mol of ( MnO4-) times 5??
Where is the 5 come from?
What is the relationship between (S2O3 2- ) and ( MnO4-)?
Please help me, thanks.

 

[symbolipoint]

Look at each half-reaction! What is necessary to balance the electrons?

 

[symbolipoint]

My thought may have been off-track a little bit. Maybe you do have the reactions that are needed, but the titration is intended to find the excess iodine remaining from the first reaction.

 

[Borek]

My thought may have been off-track a little bit. Maybe you do have the reactions that are needed, but the titration is intended to find the excess iodine remaining from the first reaction.

It is still about following electrons - we use iodine as an intermediate to make the chemistry efficient, but in the end thiosulfate is oxidized by permanganate. This can be written as a single, hypothetical reaction and then the ratio between thiosulfate and permanganate is obvious. The problem is, such reaction doesn't have properties we expect from reactions used in titration - it is either not fast enough, nor specific enough (side reactions are often a problem in such cases).

 

[mimi88]

Thank guys, now it makes more sense to me:)