Question

How many milliliters of a 0.211 M HI solution are needed to reduce 24.0 mL of a 0.354 M KMnO4 solution according to the following equation:
12HI + 2KMnO4 + 2H2SO4 → 6I2 + Mn2SO4 + K2SO4 + 8H2O

Please explain with steps. I want to understand how to solve the problem, and I don't want the answer simply handed to me with no explanation. Thanks in advance! :)

Answer 1

Answer:

The answer to your question is 242 ml

Explanation:

Data

HI 0.211 M   Volume = x

KMnO₄ 0.354 M   Volume = 24 ml

Balanced Chemical reaction

     12HI + 2KMnO₄ + 2H₂SO₄ → 6I₂ + Mn₂SO₄ + K₂SO₄ + 8H₂O

Process

1.- Calculate the moles of KMnO₄  0.354 M in 24 ml

Molarity = moles / volume (L)

moles = Molarity x volume (L)

moles = 0.354 x 0.024

moles = 0.0085

2.- From the balanced chemical reaction we know that HI and KMnO₄ react in the proportion 12 to 2. Then,

              12 moles of HI --------------- 2 moles of KMnO₄

                x                     --------------- 0.0085 moles of KMnO₄

             x = (0.0085 x 12)/2

             x = 0.051 moles of HI

3.- Calculate the milliliters of HI 0.211 M

Molarity = moles/volume

Volume = moles/molarity

Volume = 0.051/0.211

Volume = 0.242 L or Volume = 242 ml