The volume of oxygen required for the complete oxidation of 160 liters of sulfur (IV) oxide to sulfur (VI) oxide.

The volume is determined by the formula: V = Vn n, where Vn is the molar volume equal to 22.4 l / mol, and n is the amount of substance.
Let’s calculate the amount of sulfur oxide (IV) substance according to the formula.
Let’s compose the reaction equation, find the quantitative ratios of sulfur (IV) oxide and oxygen. From the reaction equation, it follows that 2 mol of oxide and 1 mol of oxygen enter the reaction, that is, they react 2: 1, so the amount of oxygen substance will be 2 times less than the amount of sulfur oxide substance. n (oxygen) = 7.14: 2 = 3.57 mol.
Knowing the amount of oxygen, you can find its volume: V = 2.24 l / mol x 3.57 mol = 79.97 l.