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 13.8 mL
of the KMnO4
solution?
To solve this problem, we can use the following balanced redox reaction between hydrogen peroxide (H2O2) and potassium permanganate (KMnO4):
5 H2O2 + 2 KMnO4 + 3 H2SO4 → 5 O2 + 2 MnSO4 + 8 H2O + K2SO4
From the balanced equation, we can see that 5 moles of H2O2 react with 2 moles of KMnO4. Therefore, the molar ratio is 5:2.
First, calculate the number of moles of KMnO4 that reacted:
Moles KMnO4 = Molarity KMnO4 x Volume KMnO4
Moles KMnO4 = 1.68 M x 0.0138 L
Moles KMnO4 = 0.023184 moles
Next, using the molar ratio from the balanced equation, calculate the moles of H2O2 that reacted:
Moles H2O2 = (2/5) x Moles KMnO4
Moles H2O2 = (2/5) x 0.023184
Moles H2O2 = 0.0092736 moles
Now, calculate the mass of H2O2 in grams:
Molar mass H2O2 = 34.0147 g/mol
Mass H2O2 = Moles H2O2 x Molar mass H2O2
Mass H2O2 = 0.0092736 moles x 34.0147 g/mol
Mass H2O2 = 0.3158 grams
Therefore, approximately 0.3158 grams of hydrogen peroxide (H2O2) was dissolved in 100 mL of water.