The balanced equation for sulfur dioxide formation is:S s + O2 g SO2 g First, we need to determine the moles of each reactant. The molar mass of sulfur S is 32.07 g/mol, and the molar mass of oxygen O2 is 32.00 g/mol.Moles of sulfur = 10 g / 32.07 g/mol = 0.312 molesMoles of oxygen = 15 g / 32.00 g/mol = 0.469 molesNow, we need to find the limiting reactant. The stoichiometry of the balanced equation shows that 1 mole of sulfur reacts with 1 mole of oxygen. Therefore, we can compare the mole ratios:Mole ratio of sulfur to oxygen = 0.312 moles / 0.469 moles = 0.665Since the mole ratio is less than 1, sulfur is the limiting reactant. Now, we can calculate the amount of sulfur dioxide that can be produced:Moles of SO2 = moles of limiting reactant sulfur = 0.312 molesThe molar mass of sulfur dioxide SO2 is 64.07 g/mol.Mass of SO2 = moles of SO2 molar mass of SO2 = 0.312 moles 64.07 g/mol = 19.99 gSo, the student can produce 19.99 grams of sulfur dioxide.To find the amount of excess reactant oxygen , we can use the stoichiometry of the balanced equation:Moles of oxygen used = moles of sulfur = 0.312 molesExcess moles of oxygen = initial moles of oxygen - moles of oxygen used = 0.469 moles - 0.312 moles = 0.157 molesMass of excess oxygen = moles of excess oxygen molar mass of oxygen = 0.157 moles 32.00 g/mol = 5.02 gThe limiting reactant is sulfur, and there are 5.02 grams of excess oxygen.