How many milliliter of a 0.206M HI solution are needed to reduce 22.5 mL of a 0.374 M KMnO4 solution according to the following equation
10HI+2KMNO4+3H2SO4- 5I2 + 2MnSO4+8H2O

204mL

Explanation:

We'll begin by calculating the number of mole of KMnO4 in 22.5 mL of 0.374 M KMnO4. This is illustrated below:

Molarity of KMnO4 = 0.374 M

Volume = 22.5 mL = 22.5/1000 = 2.25x10^-2L

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole of KMnO4 = 0.374x2.25x10^-2

Mole of KMnO4 = 8.415x10^-3 mole

The equation for the reaction is given below:

10HI + 2KMnO4 + 3H2SO4 —> 5I2 + K2SO4 + 2MnSO4 + 8H2O

From the balanced equation above,

2 moles of KMnO4 reacted with 10 moles of HI.

Therefore, 8.415x10^-3 mole of KMnO4 will react with = (10x8.415x10^-3) /2 = 0.0421 mole of HI.

Now, with this amount ( i.e 0.0421 mole) of HI, we can obtain the volume of 0.206M HI solution needed for the reaction. This is illustrated below:

Molarity of HI = 0.206M

Mole of HI = 0.0421 mole

Volume =?

Molarity = mole /Volume

Volume = mole/Molarity

Volume = 0.0421/0.206

Volume = 0.204L

Now let us convert 0.204L to mL. This is illustrated below:

1L = 1000mL

0.204L = 0.204 x 1000 = 204mL

Therefore, 204mL of 0.206M HI solution is needed for the reaction.

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