Question

How many milliliters of a 0.212 M HI solution are needed to reduce 20.5 mL of a 0.358 M KMnO4 solution according to the following equation: 10HI + 2KMnO4 + 3H2SO4 → 5I2 + 2MnSO4 + K2SO4 + 8H2O

Answer 1

Volume = 0.17 L

Step-by-step explanation:

`Molarity=(Moles\ of\ solute)/(Volume\ of\ the\ solution)`

`Moles =Molarity * {Volume\ of\ the\ solution}`

For
`KMnO_4` :

Molarity = 0.358 M

Volume = 20.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.5×10⁻³ L

Moles of
`KMnO_4` :

`Moles =0.358 * {20.5* 10^(-3)}\ moles=0.007339\ moles`

According to the reaction shown below:-

`10HI + 2KMnO_4 + 3H_2SO_4\rightarrow 5I_2 + 2MnSO_4 + K_2SO_4 + 8H_2O`

2 moles of
`KMnO_4` reacts with 10 moles of HI

Also,

1 mole of
`KMnO_4` reacts with 5 moles of HI

So,

0.007339 mole of
`KMnO_4` reacts with 5*0.007339 moles of HI

Moles of HI = 0.036695 moles

Volume = ?

Molarity = 0.212 M

So,

`0.212=(0.036695)/(Volume\ of\ the\ solution)`

Volume = 0.17 L