Asked by Kayla
How many milliliters of a 0.200 M HI solution are needed to reduce 21.5 mL of a 0.365 M KMnO4 solution according to the following equation:
10HI+2KMnO4+3H2SO4 ---> 5I2+ 2MnSO4 +K2SO4+ 8H2O
10HI+2KMnO4+3H2SO4 ---> 5I2+ 2MnSO4 +K2SO4+ 8H2O
Answers
Answered by
DrBob222
mols KMnO4 = M x L = ?
Mole HI = 5x that (from the coefficients of 10/2)
M HI = mols/L. You know M and mols, solve for L and convert to mL.
Mole HI = 5x that (from the coefficients of 10/2)
M HI = mols/L. You know M and mols, solve for L and convert to mL.