To determine the limiting reactant and the amount of sulfur dioxide produced, we need to first find the moles of each reactant and then compare their mole ratios.The molar mass of sulfur S is 32.06 g/mol, and the molar mass of oxygen O2 is 32.00 g/mol.For sulfur:moles of S = mass / molar mass = 25 g / 32.06 g/mol = 0.779 molFor oxygen:moles of O2 = mass / molar mass = 30 g / 32.00 g/mol = 0.938 molNow, we need to compare the mole ratios of the reactants to the balanced chemical equation. The balanced equation shows that 1 mole of S reacts with 1 mole of O2 to produce 1 mole of SO2.Mole ratio of S to O2 = 0.779 mol / 0.938 mol = 0.831Since the mole ratio is less than 1, sulfur is the limiting reactant.Now, we can calculate the amount of sulfur dioxide produced using the stoichiometry of the balanced equation:1 mole of S produces 1 mole of SO20.779 mol of S will produce 0.779 mol of SO2To find the mass of SO2 produced, we need to multiply the moles of SO2 by its molar mass 64.06 g/mol :mass of SO2 = moles of SO2 molar mass = 0.779 mol 64.06 g/mol = 49.9 gTherefore, the limiting reactant is sulfur, and 49.9 g of sulfur dioxide will be produced.