Question

A 150.0 mL solution of 3.241 M strontium nitrate is mixed with 205.0 mL of a 3.569 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. mass: g Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration. [Na+]= M [NO−3]= M [Sr2+]= M [F−]=

Answer 1

Sr (NO₂) + 2NaF → Sefa + 2 Na Noz I 3.241 X 150 1000 3.569 8205 = 0.486 mol = 100 0.732 mol -0.486 mol - 0.486 mol +0.486 mol 2x0.486 mol Ε 0.246 mol 0.486 mol 0.972mol . Amount of SrF formed = 0.486 mol Mass of SFF₂ = Number of moles X Molar mass of Sofa = 0.486 mol x 125.62 g/mol = 61.05 g. [Se 2+] = 0.486 mol x 1000 = 1.37 M 355 (NAF) (58F2). Cr-] = [0.246 + 2x0.486) x1000 = 3.43 m 355 Na+] = (NGF) (NaNO₂) 0.46 + 0.972 x 1000 355 = 3.437 [no] = 0.972 x 1000 355 = 2.747