(b) How many moles of H 2 O 2 were present in the 19.4-g sample of bleach?The ac

Question

(b) How many moles of H2O2 were present in the 19.4-g sample of bleach?The active agent in many hair bleaches is hydrogen peroxide. The amount of H2O2 in 19.4 g of hair bleach was determined by titration with a standard potassium permanganate solution.

2 MnO4(aq) + 5 H2O2(aq) + 6 H+(aq) 5 O2(g) + 2 Mn2+(aq) + 8 H2O(l)

(a) How many moles of MnO4 were required for the titration if 44.9 mL of 0.107 M KMnO4 was needed to reach the end point?

(b) How many moles of H2O2 were present in the 19.4-g sample of bleach?

(c) How many grams of H2O2 were in the sample?

(d) What is the mass percent of H2O2 in the sample?

Explanation / Answer

Given :

2 MnO4(aq) + 5 H2O2(aq) + 6 H+(aq) 5 O2(g) + 2 Mn2+(aq) + 8 H2O(l)

a). Calculation of moles of MnO4 :

Given :

[KMnO4]= 0.107 g, volume in mL = 44.9 mL

Calculation of moles of KMnO4

Mol KMnO4 = Volume in   L x Molarity of KMnO4

= 0.0449 L x 0.107 M

= 0.00480 mol

From we know moles of KMnO4 = moles of MnO4-

Moles of MnO4- = 0.00480 moles were required

b). Calculation of moles of H2O2

We use mol ratio to calculate moles H2O2

Mole ratio of H2O2 : MnO4- is 5 : 2

Moles of H2O2 = moles of MnO4- x 5 mol H2O2 / 2 mol MnO4-

= 0.00480 mol MnO4- x 5 mol H2O2 / 2 mol MnO4-

=0.0120 mol H2O2

c). Calculation of gram of H2O2 in the sample

Mass of H2O2 = Number of moles of H2O2 x molar mass of H2O2

= 0.0120 mol H2O2 x 34.0147 g /mol

= 0.4085 g H2O2

d). Mass percent of H2O2 in the sample = (Mass of H2O2 in g / Mass of sample )x 100

= 0.4085 g / 19.4 x 100

= 2.1 %

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