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

In summary, the conversation discusses the ratio of KMnO4 to I2 and the ratio of I2 to S2O3 2-, and the relationship between these substances. The mole of S2O3 2- is equal to the mole of MnO4- times 5, which is necessary to balance the electrons. The titration is intended to find the excess iodine remaining from the first reaction, and in the end, thiosulfate is oxidized by permanganate with a specific and efficient reaction.
  • #1
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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.
 
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  • #2
Look at each half-reaction! What is necessary to balance the electrons?
 
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  • #3
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.
 
  • #4
symbolipoint said:
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).
 
  • #5
Thank guys, now it makes more sense to me:)
 

1. Why are moles of (S2O3 2-) equal to mol of ( MnO4-) times 5?

This is because the balanced chemical equation for the reaction between (S2O3 2-) and ( MnO4-) has a 1:5 mole ratio. This means that for every 1 mole of ( MnO4-), 5 moles of (S2O3 2-) are needed for the reaction to occur. Therefore, the number of moles of (S2O3 2-) is always equal to the number of moles of ( MnO4-) times 5.

2. How do you determine the mole ratio between (S2O3 2-) and ( MnO4-)?

The mole ratio between (S2O3 2-) and ( MnO4-) can be determined from the balanced chemical equation for the reaction between the two substances. The coefficients of the two compounds in the equation represent the mole ratio. In this case, the coefficient of (S2O3 2-) is 5, indicating that the mole ratio between (S2O3 2-) and ( MnO4-) is 5:1.

3. Is the mole ratio between (S2O3 2-) and ( MnO4-) always 5:1?

Yes, the mole ratio between (S2O3 2-) and ( MnO4-) is always 5:1 as long as the chemical reaction remains the same. This is because the coefficients in a balanced chemical equation represent the relative number of moles of each substance involved in the reaction.

4. Can the number of moles of (S2O3 2-) ever be greater than the number of moles of ( MnO4-)?

No, the number of moles of (S2O3 2-) can never be greater than the number of moles of ( MnO4-) in this reaction. This is because the mole ratio between the two substances is 5:1, meaning that for every 1 mole of ( MnO4-), there must be 5 moles of (S2O3 2-) present for the reaction to occur.

5. Why is it important to have the correct mole ratio in a chemical reaction?

Having the correct mole ratio in a chemical reaction is important because it ensures that the reaction proceeds efficiently and all of the reactants are consumed. If the mole ratio is not correct, there may be excess reactants or incomplete reaction, leading to lower yield and potentially affecting the quality of the final product. Additionally, the mole ratio is needed to accurately calculate the amount of products produced in the reaction.

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