In the reaction between potassium dichromate (K2Cr2O7) and hydrogen peroxide (H2O2) in an acidic solution, balance the equation and calculate the volume of 0.2 M potassium dichromate required to completely oxidize 25.0 mL of 0.1 M hydrogen peroxide to water and oxygen gas.
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The balanced equation for the reaction between potassium dichromate (K2Cr2O7) and hydrogen peroxide (H2O2) in an acidic solution is as follows:
K2Cr2O7 + 8H2SO4 + 3H2O2 → K2SO4 + Cr2(SO4)3 + 8H2O + 3O2
To calculate the volume of 0.2 M potassium dichromate required to completely oxidize 25.0 mL of 0.1 M hydrogen peroxide, we need to determine the stoichiometry of the reaction.
From the balanced equation, we can see that the ratio between potassium dichromate and hydrogen peroxide is 1:3. Therefore, for every 1 mole of hydrogen peroxide, we need 1/3 moles of potassium dichromate.
To convert from moles to volume, we can use the formula:
concentration (M) = moles/volume (L)
Rearranging the formula, we have:
volume (L) = moles/concentration (M)
First, let's calculate the moles of hydrogen peroxide:
moles of H2O2 = concentration × volume
= 0.1 M × 0.025 L
= 0.0025 mol
Since the ratio is 1:3, the moles of potassium dichromate required will be 1/3 × 0.0025 mol = 0.000833 mol.
Now, let's calculate the volume of 0.2 M potassium dichromate required:
volume (L) = moles/concentration
= 0.000833 mol / 0.2 M
= 0.00417 L
= 4.17 mL
Therefore, 4.17 mL of 0.2 M potassium dichromate is required to completely oxidize 25.0 mL of 0.1 M hydrogen peroxide to water and oxygen gas.
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