What mass of excess reactant remains at the end of the reaction if 90.0 g of SO2 are mixed with 100.0 g of O2? 2SO₂ + O₂ → 2SO₃

Question

Quizplus · Accepted Answer

First, calculate the moles of SO2 and O2. SO2: 90.0 g / 64.07 g/mol = 1.404 mol. O2: 100.0 g / 32.00 g/mol = 3.125 mol. The reaction requires 1 mol of O2 for every 2 mol of SO2. Therefore, 1.404 mol of SO2 requires 0.702 mol of O2. Since there is 3.125 mol of O2 available, O2 is in excess. The excess O2 is 3.125 mol - 0.702 mol = 2.423 mol. Convert this to grams: 2.423 mol * 32.00 g/mol = 77.5 g.

What Mass of Excess Reactant Remains at the End of the Reaction