What volume of oxygen is required for the oxidation of 0.6 mol of sulfur oxide (IV)?

SO2 + 0.5O2 + SO3
the number of moles of oxygen is two times less than the number of moles of sulfur dioxide, which means that the volume of oxygen will be equal to (0.6 / 2) * 22.4 = 6.72 liters.

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