text: Your two basic algebraic equations are:\nmoles Na + moles K = moles Cl \n(g Na+) + (g K+) + (g Cl-) = 6.4100 g\n\nThe mass of the final precipitate means that there are 7.50921 x 10-2 mol of chloride ions present in the initial mixture.\nSee:\nmol Cl = mol AgCl = g AgCl/MW AgCl = 12.7559/169.87 = 7.50921 x 10 -2 mol\n\nThe mass of the chloride component of the starting materials is then determined by:\n\ng Cl = MW Cl x mol Cl = (7.50921 x 10-2 mol) x (35.45 g/mol) = 2.662 g\n\nPlug that back into the second eqn and you get:\n\n(g Na+) + (g K+) + (2.662 g) = 6.4100 g\nor\n(g Na+) + (g K+) = 3.748 g\n\nConvert back to moles and plug that into the first equation and you get:\n\nmoles Na + moles K = 7.50921 x 10 -2 mol\nor\nmoles Na + ((g K+) / (39.1 g/mol)) = 7.50921 x 10 -2 mol\nor\n((g Na+) / (22.99 g/mol)) + ((3.748 - g Na+) / (39.1 g/mol)) = 7.50921 x 10 -2 mol\n\nAnd now solve for g Na + \n\nDivide that by 3.748 g and you're there!