Answer:
0,508g of Hâ‚‚Oâ‚‚
Explanation:
For the reaction:
2KMnO₄(aq) + H₂O₂(aq) + 3H₂SO₄(aq) → 3O₂(g) + 2MnSO₄(aq) + K₂SO₄(aq) + 4H₂O(l)
2 moles of KMnOâ‚„ react with 1 mol of Hâ‚‚Oâ‚‚.
In the titration, moles of KMnOâ‚„ required were:
1,68M×0,0178L = 0,0299 moles of KMnO₄. Moles of H₂O₂ are:
0,0299 moles of KMnO₄×[tex]\frac{1molH_{2}O_{2}}{2molKMnO_{4}}[/tex] = 0,01495 moles of H₂O₂. As molar mass of H₂O₂ is 34,01g/mol, mass of H₂O₂ was dissolved is:
0,01495 moles of H₂O₂×[tex]\frac{34,01g}{1molH_{2}O_{2}}[/tex] = 0,508g of H₂O₂
