Question

I'm doing a prelab right now and I cant figure out how to do this one.

What volume of 0.0 10 M potassium permanganate solution would be required to completely react with 0.50 g of hydrogen peroxide solution. The commercial solution is 3 % by mass of pure H2O2 in an aqueous solution. Calculations must be clearly shown 6.

Answer 1

Mass of H202 solution Mass percent 3 % 0.50 g We know that mass percent = mass of solute/mass of the solution * 100 mass of solute = 3 * 0.50/100 0.015 g Mass of H202= 0.015 g Number of moles of H202 mass/mol.wt. 0.015 g/(34.01 g/mol) 0.000441 mol

The balanced equation for the reaction between H202 and KMnO4 is 2 KMnO4 5 H202 +3H2SO4-- ---> 2MnSO4+ K2SO45 02+8H20 Accordingto balanced equation, 5 moles of H202 reacts with 2 moles of KMnO4 So 0.000441 mol of H202 would react with 0.000441 *2/5 0.000176 mol of KMnO4 Number of moles of KMnO4 0.000176 mol Volume of KMnO4= moles/molarity 0.000176 mol/ 0.010 M 0.017642 L 17.64187 mL Volume of KMnO4 17.64 mL