Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. for example, a solution of hydrogen peroxide, h2o2, can be titrated against a solution of potassium permanganate, kmno4. the following equation represents the reaction: 2kmno4(aq) h2o2(aq) 3h2so4(aq)→3o2(g) 2mnso4(aq) k2so4(aq) 4h2o(l) a certain amount of hydrogen peroxide was dissolved in 100. ml of water and then titrated with 1.68 m kmno4. what mass of h2o2 was dissolved if the titration required 20.8 ml of the kmno4 solution

0.595 g of H₂O₂ were dissolved.

Explanation:

1) Balanced chemical equation of the titration (given):

2KMnO₄(aq) + H₂O₂(aq) + 3H₂SO₄(aq) →

                                                → 3O₂(g) + 2MnSO₄(aq)+  K₂SO₄(aq) + 4H₂O(l)

2) Mole ratios of KMnO₄(aq) and H₂O₂(aq) (the reactants the question deals with):

2 moles 2KMnO₄(aq) : 1 mol H₂O₂(aq)

3) Number of moles of KMnO₄ used:

V = 20.8 mL = 0.0208 literM = 1.68 MM = n of solute / V of solution in liter ⇒ n of KMnO₄ = M × V = 1.68 M × 0.0208 liter = 0.0349 moles

4) Number of moles of H₂O₂ used:

Set the proportion using the stoichimetric ratio and the actual number of moles of KMnO₄:

       2 moles KMnO₄ / 1 mol H₂O₂ = 0.0349 moles KMnO₄ / x

       ⇒ x = 0.0349 / 2 moles H₂O₂ = 0.0175 moles H₂O₂

5) Mass of H₂O₂ dissolved:

molar mass H₂O₂ = 2(1.008g/mol) + 2(15.999 g/mol) = 34.014 g/molmolar mass = mass in grams / number of moles ⇒ ⇒ mass in grams = molar mass × number of moles = 34.01 g/mol × 0.0175 moles = 0.595 g.

So, the answer, using the appropiate number of signficant digits (3) is 0.595 g of H₂O₂.

Ithink its called the coefficent.

This is a solution, hope i could