In the previous example, how many ml of 0.1M KMN04 are required to react with 30ml of 0.05M H2o2? The answer 3ml
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To determine the amount of KMnO4 required to react with 30 ml of 0.05M H2O2, you need to use the balanced chemical equation between KMnO4 and H2O2.
The balanced equation is:
2 KMnO4 + 3 H2O2 -> 2 MnO2 + 2 KOH + 3 H2O + 3 O2
From the balanced equation, you can see that 2 moles of KMnO4 are required to react with 3 moles of H2O2. This means the ratio of KMnO4 to H2O2 is 2:3.
Next, you need to convert the given concentration of H2O2 (0.05M) into moles. Since Molarity (M) is given in moles per liter, you need to multiply the concentration by the volume in liters:
moles of H2O2 = concentration (M) × volume (L)
moles of H2O2 = 0.05 M × 0.03 L
moles of H2O2 = 0.0015 moles
Since the ratio of KMnO4 to H2O2 is 2:3, you can set up a proportion to find the moles of KMnO4 required:
moles of KMnO4 / moles of H2O2 = 2 / 3
moles of KMnO4 = (2 / 3) × 0.0015 moles
moles of KMnO4 = 0.001 moles
Finally, you need to convert the moles of KMnO4 back into milliliters using the concentration of KMnO4 (0.1M):
volume (L) = moles / concentration (M)
volume (L) = 0.001 moles / 0.1 M
volume (L) = 0.01 L
Convert the volume from liters to milliliters:
volume (ml) = 0.01 L × 1000 ml/L
volume (ml) = 10 ml
Therefore, 10 ml of 0.1M KMnO4 is required to react with 30 ml of 0.05M H2O2.