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

Answer:

Volume = 0.17 L

Explanation:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

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 $KMnO_4$ :

$Moles =0.358 \times {20.5\times 10^{-3}}\ moles=0.007339\ moles$

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,

$0.212=\frac{0.036695}{Volume\ of\ the\ solution}$

Volume = 0.17 L