How many milliliters of a 0.264 M HI solution are needed...?

Question

...to reduce 23.3 mL of a 0.374 M KMnO4 solution according to the following equation?

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

Can I please get some help from the chemistry people here?

Answer 1

You should make the reactants and products separated as in the following equation:
10 HI + 2 KMnO4 + 3 H2SO4 ---> 5 I2 + 2 MnSO4 + K2SO4 + 8 H2O
23.3 mL of a 0.374 M KMnO4 contains 0.0233*0.374 = 0.008714 Mol of KMnO4. Since 5 portion of HI is required to react with 1 portion of KMnO4 as shown in the equation, 5*0.008714 = 0.04357 Mol of HI required.
(0.04357 Mol) /(0.264 Mol/L) = 0.165L = 165mL
165mL of a 0.264 M HI solution are needed.