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.
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?
1 answer