To determine the limiting reactant and the amount of magnesium oxide that can be produced, we need to use the stoichiometry of the balanced chemical equation for the reaction:2 Mg s + O g 2 MgO s First, we need to find the molar mass of magnesium Mg and oxygen O .Molar mass of Mg = 24.31 g/molMolar mass of O = 2 16.00 g/mol = 32.00 g/molNow, we'll convert the given masses of Mg and O to moles.Moles of Mg = 10 g / 24.31 g/mol = 0.411 molesMoles of O = 15 g / 32.00 g/mol = 0.469 molesNext, we'll determine the mole ratio of Mg to O in the balanced chemical equation:Mole ratio Mg:O = 2:1Now, we'll compare the mole ratio of the given reactants to the balanced chemical equation:Mole ratio of given reactants Mg:O = 0.411 moles / 0.469 moles = 0.876Since the mole ratio of the given reactants 0.876 is greater than the mole ratio in the balanced chemical equation 2:1 = 0.5 , this means that magnesium Mg is the limiting reactant.Now, we'll find out how much magnesium oxide MgO can be produced using the limiting reactant Mg . We'll use the stoichiometry of the balanced chemical equation:2 moles of Mg 2 moles of MgO0.411 moles of Mg x moles of MgOx = 0.411 moles of Mg 2 moles of MgO / 2 moles of Mg = 0.411 moles of MgONow, we'll convert moles of MgO to grams:Mass of MgO = 0.411 moles 40.31 g/mol = 16.55 gTherefore, the limiting reactant is magnesium Mg , and 16.55 grams of magnesium oxide MgO can be produced.