20ml solution containing 0.2g of an impure sample of h2o2 reacts with 0.316g of kmno4 in the presence of h2so4. Find the purity of h2o2 in the mixture.
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To find the purity of H2O2 in the mixture, we can use the stoichiometry of the reaction between H2O2 and KMnO4. The balanced equation for the reaction is:
5H2O2 + 2KMnO4 + 3H2SO4 -> 5O2 + 2MnSO4 + 5H2O + K2SO4
From the balanced equation, we can see that 5 moles of H2O2 react with 2 moles of KMnO4.
First, let's calculate the number of moles of KMnO4 used:
Molar mass of KMnO4 = 39.10g/mol (K) + 54.94g/mol (Mn) + 4 * 16.00g/mol (O) = 158.04g/mol
Number of moles of KMnO4 = 0.316g / 158.04g/mol = 0.002 mol
According to the stoichiometry, 1 mole of H2O2 reacts with 0.4 moles of KMnO4:
Number of moles of H2O2 = 0.002 mol / 0.4 = 0.005 mol
Now, let's calculate the number of moles of H2O2 in the impure sample:
Molar mass of H2O2 = 1.01g/mol (H) + 1.01g/mol (H) + 16.00g/mol (O) = 34.02g/mol
Number of moles of H2O2 in the impure sample = 0.2g / 34.02g/mol = 0.0059 mol
The purity of H2O2 in the mixture can be calculated as:
Purity of H2O2 = (Number of moles of H2O2 in the impure sample / Number of moles of H2O2 calculated) * 100
Purity of H2O2 = (0.0059 mol / 0.005 mol) * 100 = 118%
Therefore, the purity of H2O2 in the mixture is 118%.