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.

the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation: