Final answer:
To react completely with 0.870 L of oxygen gas, 1.74 L of sulfur dioxide gas is required based on their stoichiometric ratio in the reaction to form sulfur trioxide.
Step-by-step explanation:
The question asks about the volume of sulfur dioxide gas that will react completely with a given volume of oxygen gas based on the stoichiometry of the balanced chemical equation for the reaction between sulfur dioxide and oxygen to form sulfur trioxide.
The balanced chemical equation is:
2 SO2(g) + O2(g) → 2 SO3(g)
According to the equation, two moles of SO2 react with one mole of O2 to produce two moles of SO3. Since the conditions of temperature and pressure are constant, we can use the direct relationship of the volumes of gases involved in the reaction (Avogadro's Law). This means that the ratio of the volumes of gases will be the same as their mole ratio in the balanced chemical equation.
Given that 0.870 L of O2 participates in the reaction, and the mole ratio of SO2 to O2 is 2:1, the volume of SO2 needed would be twice the volume of O2. Therefore, the volume of SO2 gas required to react with 0.870 L of O2 will be:
2 × 0.870 L = 1.74 L
So, the correct answer is (a) 1.74 L.