You are given a 1.50-g mixture of sodium nitrate and sodium chloride. You dissolve this mixture into 100 mL of water and then add an excess of 0.500 M silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of 0.641 g. a. If you had an extremely magnified view of the solution (to the atomic- molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the mass percent of sodium chloride in the original unknown mixture.

When the sodium nitrate (NaNO₃) and sodium chloride (NaCl) mixture is dissolved in water, we would observe the following species, including their charges: 1. Na⁺ (aq) - sodium ions 2. NO₃⁻ (aq) - nitrate ions 3. Cl⁻ (aq) - chloride ions Additionally, once we add the silver nitrate (AgNO₃) solution, we would also observe: 4. Ag⁺ (aq) - silver ions

The reaction that takes place is between silver ions (Ag⁺) and chloride ions (Cl⁻) to form an insoluble precipitate, silver chloride (AgCl). The balanced net ionic equation, including phases and charges, is: \( \newline Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s) \newline \)

First, we need to calculate the number of moles of silver chloride formed using its mass: Mass of AgCl = 0.641 g Molar mass of AgCl = 107.87 g/mol (Ag) + 35.45 g/mol (Cl) = 143.32 g/mol Number of moles of AgCl = (mass of AgCl)/(molar mass of AgCl) Number of moles of AgCl = (0.641 g)/(143.32 g/mol) = 0.00447 mol Since one mole of silver chloride is formed per mole of chloride ion, the number of moles of sodium chloride is equal to that of silver chloride: Number of moles of NaCl = 0.00447 mol Now we can calculate the mass of sodium chloride originally present in the mixture: Molar mass of NaCl = 22.99 g/mol (Na) + 35.45 g/mol (Cl) = 58.44 g/mol Mass of NaCl = (number of moles of NaCl) × (molar mass of NaCl) Mass of NaCl = (0.00447 mol) × (58.44 g/mol) = 0.261 g Finally, we can find the mass percent of sodium chloride in the original mixture: Mass percent of NaCl = (mass of NaCl)/(total mass of the mixture) × 100 Mass percent of NaCl = (0.261 g)/(1.50 g) × 100 = 17.4 % So, the mass percent of sodium chloride in the original unknown mixture is 17.4%.