Hydrogen peroxide, H2O2, solution can be purchased in drug stores and used as an antiseptic. A sample of such a solution weighing 1.00 grams was acidified with sulfuric acid and titrated with .02000 M solution of KMnO4. The unbalanced reaction is H2O2(aq) + MnO4-(aq)--> O2(g)+Mn=2(aq)
The titration required 17.60ml of KMnO4 solution. What is the mass percent of H2O2 in the original solution? (balance the equation first.)
H2O2(aq) + 2H+(aq) + 2MnO4-(aq) --> 2H2O(l) + 2Mn2+(aq) + O2(g)
Mass percent of H2O2 = (1.00 g H2O2/1.00 g solution) x 100% = 100%
To find the mass percent of H2O2 in the original solution, we first need to balance the equation:
H2O2(aq) + MnO4-(aq) --> O2(g) + Mn2+(aq)
Now, let's calculate the moles of KMnO4 used in the titration:
Molarity of KMnO4 solution = 0.02000 M
Volume of KMnO4 solution used = 17.60 mL = 0.01760 L
Moles of KMnO4 = Molarity × Volume
= 0.02000 M × 0.01760 L
= 0.000352 mol
Using the balanced equation, we can see that the ratio between H2O2 and KMnO4 is 1:5. This means that 1 mole of H2O2 reacts with 5 moles of KMnO4.
Therefore, the moles of H2O2 in the original solution are:
Moles of H2O2 = 0.000352 mol × (1/5)
= 0.0000704 mol
Next, let's find the molar mass of H2O2:
Molar mass of H2O2 = (2 × Atomic mass of Hydrogen) + (2 × Atomic mass of Oxygen)
= (2 × 1.008 g/mol) + (2 × 16.00 g/mol)
= 2.016 g/mol + 32.00 g/mol
= 34.016 g/mol
Finally, let's calculate the mass percent of H2O2 in the original solution:
Mass of H2O2 = Moles of H2O2 × Molar mass of H2O2
= 0.0000704 mol × 34.016 g/mol
= 2.374144 g
Mass percent of H2O2 = (Mass of H2O2 / Mass of solution) × 100
= (2.374144 g / 1.00 g) × 100
= 237.4144%
Therefore, the mass percent of H2O2 in the original solution is approximately 237.41%.
To find the mass percent of H2O2 in the original solution, we need to determine the number of moles of KMnO4 used in the titration.
The balanced equation for the reaction is:
5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 5 O2(g) + 2 Mn2+(aq) + 8 H2O(l)
From the balanced equation, we can see that the stoichiometric ratio between H2O2 and KMnO4 is 5:2.
Given that the volume of KMnO4 solution used in the titration is 17.60 mL and the concentration of KMnO4 is 0.02000 M, we can calculate the number of moles of KMnO4 used:
moles of KMnO4 = concentration of KMnO4 × volume of KMnO4 solution
= 0.02000 M × 0.01760 L
= 0.000352 mol
Since the stoichiometric ratio between H2O2 and KMnO4 is 5:2, the number of moles of H2O2 in the original solution is:
moles of H2O2 = (5/2) × moles of KMnO4
= (5/2) × 0.000352 mol
= 0.000880 mol
To determine the mass of H2O2 in the original solution, we need to use the molar mass of H2O2, which is 34.0147 g/mol.
mass of H2O2 = moles of H2O2 × molar mass of H2O2
= 0.000880 mol × 34.0147 g/mol
= 0.02991176 g
Finally, to calculate the mass percent of H2O2 in the original solution, we divide the mass of H2O2 by the mass of the original solution and multiply by 100:
mass percent of H2O2 = (mass of H2O2 / mass of original solution) × 100
= (0.02991176 g / 1.00 g) × 100
= 2.991176%
Therefore, the mass percent of H2O2 in the original solution is approximately 2.991176%.