When silver nitrate is added to the mixture, it reacts with sodium chloride to form silver chloride precipitate. The reaction is as follows:AgNO3 + NaCl -> AgCl + NaNO3The sodium nitrate does not react with silver nitrate and remains in the solution. The mass of the precipitate 20.4 g is the mass of silver chloride formed. We can use the stoichiometry of the reaction to determine the mass of sodium chloride in the original mixture.First, we need to find the molar mass of each compound:- Sodium chloride NaCl : 58.44 g/mol- Silver chloride AgCl : 143.32 g/molNow, we can use the mass of the precipitate AgCl to find the moles of AgCl formed:moles of AgCl = mass of AgCl / molar mass of AgClmoles of AgCl = 20.4 g / 143.32 g/mol = 0.1423 molSince the stoichiometry of the reaction is 1:1, the moles of NaCl in the original mixture are equal to the moles of AgCl formed:moles of NaCl = 0.1423 molNow, we can find the mass of NaCl in the original mixture:mass of NaCl = moles of NaCl * molar mass of NaClmass of NaCl = 0.1423 mol * 58.44 g/mol = 8.32 gSince the total mass of the mixture is 15.0 g, the mass of sodium nitrate in the mixture can be found by subtracting the mass of sodium chloride:mass of NaNO3 = total mass - mass of NaClmass of NaNO3 = 15.0 g - 8.32 g = 6.68 gFinally, we can find the mass percentage of each component in the mixture:mass percentage of NaCl = mass of NaCl / total mass * 100mass percentage of NaCl = 8.32 g / 15.0 g * 100 = 55.47%mass percentage of NaNO3 = mass of NaNO3 / total mass * 100mass percentage of NaNO3 = 6.68 g / 15.0 g * 100 = 44.53%So, the mass percentage of sodium chloride in the mixture is 55.47%, and the mass percentage of sodium nitrate is 44.53%.