The mass of precipitate formed when 50 mL of 16.9% aqueous solution of AgNO ₃ is mixed with 50 mL of 5.8% NaCl solution is------------g [Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5]

Step-by-step solution

Mass of AgNO ₃ in 50 mL of 16.9% solution = 16.9 100 50 = 8.45 g Molar mass of AgNO ₃ = 107.8 + 14 + 3 16 = 169.8 g/mol Moles of AgNO ₃ = 8.45 169.8 0.0497 mol Mass of NaCl in 50 mL of 5.8% solution = 5.8 100 50 = 2.9 g Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol Moles of NaCl = 2.9 58.5 0.0495 mol The balanced chemical equation is: AgNO ₃ + NaCl AgCl + NaNO ₃ Since 1 mole of AgNO ₃ reacts with 1 mole of NaCl, NaCl is the limiting reagent. Moles of AgCl precipitate formed = 0.0495 mol Molar mass of AgCl = 107.8 + 3