Question

What mass of Na, CrO4 is required to precipitate all of the silver ions from 35.0 mL of a 0.200 M solution of AgNO3? Mass = g

Answer 1

1 + 2 AgNO3 lag) Aq, Cr 04 (s) + 2NaNO3 (99) Na2 Croy Cagl + 2 AgNO3 lag molar man of Na2 CrOy = te 161.97 g/mol volume of AgNO3 Solution : 35.0mL = 0.035 L 1: 1000 ML = IL) concentration of AgNO3 solution = 0.200M moles of AgNO3 = concentration (m) X volumell) = 0.200 M X 0.035 L = 0.007 mol 2 moles of AgNO3 react with I mole of Na2Croy 0.007 mol of AgNO3 react with moles of Na2Croy = = 1 mol X0.007 mol = 0.0035 mol mol moles of N92C704 require to precipitate all Agtrony = 0.0035 mol may of Na, Croy require moles x molar mass = 0.0035 mol x 161197 g/mol - 0,567 g Na2Croy Please Rate