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
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.
To determine the volume of the HI solution needed, we can use the concept of stoichiometry.
Let's first write out and balance the equation:
10HI + 2KMnO4 + 3H2SO4 → 5I2 + 2MnSO4 + K2SO4 + 8H2O
From the balanced equation, we can see that the stoichiometric ratio between HI and KMnO4 is 10:2, which can be simplified to 5:1.
Now, we can use the following formula to calculate the volume of the HI solution needed:
(Volume of HI solution) = (Volume of KMnO4 solution) x (Molarity of KMnO4 solution) x (Stoichiometric ratio of HI and KMnO4) / (Molarity of HI solution)
Plugging in the given values:
Volume of KMnO4 solution = 21.5 mL
Molarity of KMnO4 solution = 0.365 M
Stoichiometric ratio of HI and KMnO4 = 5 (from the balanced equation)
Molarity of HI solution = 0.200 M
(Volume of HI solution) = (21.5 mL) x (0.365 M) x (5) / (0.200 M)
Now, we can calculate the volume:
(Volume of HI solution) = 398.375 mL
Therefore, you would need approximately 398.375 mL of the 0.200 M HI solution to reduce 21.5 mL of the 0.365 M KMnO4 solution.