To determine the limiting reactant, we first need to calculate the number of moles of each reactant. We can do this using the molar mass of each compound.The molar mass of sodium hydroxide NaOH is approximately:Na: 22.99 g/molO: 16.00 g/molH: 1.01 g/molTotal: 40.00 g/molThe molar mass of hydrochloric acid HCl is approximately:H: 1.01 g/molCl: 35.45 g/molTotal: 36.46 g/molNow, we can calculate the number of moles of each reactant:Moles of NaOH = 20 g / 40.00 g/mol = 0.5 molMoles of HCl = 25 g / 36.46 g/mol = 0.685 molThe balanced equation shows that the reaction occurs in a 1:1 ratio, so we can compare the moles of each reactant to determine the limiting reactant:0.5 mol NaOH : 0.685 mol HClSince there are fewer moles of NaOH, it is the limiting reactant.Now, we can determine how much product can be formed. The balanced equation shows that 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and 1 mole of H2O. Since NaOH is the limiting reactant, the amount of product formed will be based on the moles of NaOH:0.5 mol NaOH 0.5 mol NaClTo find the mass of NaCl produced, we can use the molar mass of NaCl:Na: 22.99 g/molCl: 35.45 g/molTotal: 58.44 g/molMass of NaCl = 0.5 mol 58.44 g/mol = 29.22 gTherefore, the limiting reactant is sodium hydroxide NaOH , and 29.22 grams of sodium chloride NaCl can be formed in this reaction.