First, we need to write the balanced chemical equation for the reaction:AgNO3 aq + HCl aq AgCl s + HNO3 aq From the balanced equation, we can see that the reaction occurs in a 1:1 ratio between AgNO3 and HCl.Now, we can calculate the moles of each reactant:Moles of AgNO3 = 0.1 M 0.050 L = 0.005 molesMoles of HCl = 0.2 M 0.100 L = 0.020 molesSince the reaction occurs in a 1:1 ratio, the limiting reactant is AgNO3, as there are fewer moles of it. We will use the moles of AgNO3 to determine the mass of AgCl produced.Moles of AgCl = moles of AgNO3 = 0.005 molesNow, we need to find the molar mass of AgCl:AgCl = 107.87 g/mol Ag + 35.45 g/mol Cl = 143.32 g/molFinally, we can calculate the mass of AgCl produced:Mass of AgCl = 0.005 moles 143.32 g/mol = 0.717 gSo, 0.717 grams of AgCl can be produced from the reaction.Now, let's determine the excess reactant and by how much:Since AgNO3 is the limiting reactant, HCl is in excess. We need to find out how many moles of HCl are left after the reaction.Moles of HCl used = moles of AgNO3 = 0.005 molesMoles of HCl remaining = initial moles of HCl - moles of HCl used = 0.020 moles - 0.005 moles = 0.015 molesThus, HCl is in excess by 0.015 moles.