Enter your answer in the provided box. How many milliliters of a 0.208 M HI solution...
Enter your answer in the provided box. How many milliliters of a 0.208 M HI solution are needed to reduce 20.5 mL of a 0.399 M KMnO4 solution according to the following equation: 12HI + 2KMnO4 + 2H2SO4 → 6I2 + Mn2SO4 + K2SO4 + 8H2O
Homework Answers
millimoles of KMnO4 given = 20.5 x 0.399 = 8.18
according to balanced reaction
2 millimoles KMnO4 require 12 millimoles HI
8.18 millimoles KMnO4 require 8.18 x 12 / 2 = 49.08 millimoles HI
49.08 = V x 0.208
V = 236 mL
volume of HI required = 236 mL
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