3 Formula 1 point In the preparation of sulfuric acid (H_(2)SO_(4)) sulfur dioxide must be oxidized to the less harmful sulfur trioxide. How much oxygen in liters must be available to convert 58.2 grams of sulfur dioxide according to this process? underline ( )SO_(2)+underline ( )O_(2)arrow underline ( )SO_(3)

To solve this problem, we need to follow these steps:<br /><br />1. Determine the balanced chemical equation for the reaction.<br />2. Calculate the moles of sulfur dioxide (SO2) given the mass provided.<br />3. Use the stoichiometry of the reaction to find the moles of oxygen (O2) required.<br />4. Convert the moles of oxygen to liters using the ideal gas law.<br /><br />Step 1: Balanced chemical equation<br />The balanced chemical equation for the reaction is:<br />2 SO2 + O2 → 2 SO3<br /><br />Step 2: Calculate the moles of sulfur dioxide<br />Given mass of sulfur dioxide = 58.2 grams<br />Molar mass of sulfur dioxide (SO2) = 64.06 g/mol<br />Moles of sulfur dioxide = 58.2 g / 64.06 g/mol = 0.907 mol<br /><br />Step 3: Calculate the moles of oxygen required<br />From the balanced equation, we can see that 2 moles of SO2 react with 1 mole of O2.<br />Moles of oxygen required = 0.907 mol SO2 × (1 mol O2 / 2 mol SO2) = 0.4535 mol O2<br /><br />Step 4: Convert the moles of oxygen to liters<br />At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.<br />Liters of oxygen required = 0.4535 mol O2 × 22.4 L/mol = 10.15 L<br /><br />Therefore, 10.15 liters of oxygen must be available to convert 58.2 grams of sulfur dioxide to sulfur trioxide.