150 mL sample of sulfur dioxide required 31.5 mL of 0.0100 mol/L potassium permanganate to complete the reaction. What is the concentration of sulfur dioxide in mmol
To solve this problem, we'll use the concept of stoichiometry to find the concentration of sulfur dioxide in millimoles (mmol).
From the reaction equation between sulfur dioxide (SO2) and potassium permanganate (KMnO4):
2 KMnO4 + 5 SO2 + 2 H2O -> K2SO4 + 2 MnSO4 + 2 H2SO4
we can see that the stoichiometric ratio is 5:2 between KMnO4 and SO2. This means that for every 5 moles of KMnO4, we require 2 moles of SO2.
Given that 31.5 mL of 0.0100 mol/L KMnO4 is required to react with the SO2, we can calculate the number of moles of KMnO4 used:
moles KMnO4 = concentration KMnO4 x volume KMnO4
= 0.0100 mol/L x 31.5 mL
= 0.0100 mol/L x 0.0315 L
= 0.000315 mol KMnO4
Using the stoichiometric ratio, we can determine the number of moles of SO2 reacted:
moles SO2 = (2 moles SO2 / 5 moles KMnO4) x moles KMnO4
= (2/5) x 0.000315 mol
= 0.000126 mol SO2
Finally, we can find the concentration of SO2 in mmol:
concentration SO2 = moles SO2 / volume SO2
= 0.000126 mol / 0.150 L
= 0.00084 mol/L
Converting to mmol, we multiply by 1000:
concentration SO2 = 0.00084 mol/L x 1000 mmol/mol
= 0.84 mmol/L
Therefore, the concentration of sulfur dioxide in mmol is 0.84 mmol/L.