To determine the limiting reactant and the amount of copper produced, we first need to write the balanced chemical equation for the reaction:CuO + Mg Cu + MgONext, we need to find the molar mass of each reactant:Copper oxide CuO : 63.55 g/mol Cu + 16.00 g/mol O = 79.55 g/molMagnesium Mg : 24.31 g/molNow, we can calculate the moles of each reactant:moles of CuO = 10 g / 79.55 g/mol = 0.1257 molmoles of Mg = 5 g / 24.31 g/mol = 0.2056 molNow, we need to determine the mole ratio between the reactants:mole ratio = moles of CuO / moles of Mg = 0.1257 mol / 0.2056 mol = 0.611Since the mole ratio is less than 1, CuO is the limiting reactant.Now, we can calculate the moles of copper produced using the stoichiometry of the balanced equation:moles of Cu = moles of CuO = 0.1257 molFinally, we can convert the moles of copper to grams:grams of Cu = moles of Cu molar mass of Cu = 0.1257 mol 63.55 g/mol = 7.99 gSo, the limiting reactant is copper oxide CuO , and 7.99 grams of copper can be produced.