To determine the limiting reactant and the amount of sulfur dioxide produced, we need to first look at the balanced chemical equation for the reaction:S s + O2 g SO2 g Now, we need to convert the given masses of sulfur and oxygen into moles using their respective molar masses:Molar mass of S = 32.06 g/molMolar mass of O2 = 32.00 g/molMoles of S = 5 g / 32.06 g/mol = 0.156 molesMoles of O2 = 8 g / 32.00 g/mol = 0.250 molesNext, we need to determine the mole ratio of the reactants from the balanced equation:Mole ratio S:O2 = 1:1Now, we can compare the mole ratio of the reactants to the mole ratio in the balanced equation to determine the limiting reactant:0.156 moles S / 1 = 0.1560.250 moles O2 / 1 = 0.250Since 0.156 is less than 0.250, sulfur S is the limiting reactant.Finally, we can calculate the amount of sulfur dioxide SO2 produced using the stoichiometry of the balanced equation:Moles of SO2 produced = moles of limiting reactant S = 0.156 molesNow, convert moles of SO2 to grams using the molar mass of SO2:Molar mass of SO2 = 32.06 g/mol S + 2 * 16.00 g/mol O = 64.06 g/molMass of SO2 produced = 0.156 moles * 64.06 g/mol = 9.99 gTherefore, the limiting reactant is sulfur, and 9.99 grams of sulfur dioxide can be produced.