Question

Given that 24.0 mL of 0.170 M sodium iodide reacts with 0.209 M mercury (II) nitrate solution according to the unbalanced equation Hg(NO3)2(aq) + Nal(aq) → Hgla(s) + NaNO3(aq) a) What volume of Hg(NO3)2 is required for complete precipitation of Hgl2? c) What is the mass of Hgla precipitate?

Answer 1

24 m2 0.17 M Na I mole § Na I a molasity & Volume inz = 0.17 X(24x16-3) = 0.00 408 mole Un Balance reaction Hy (NO3)2 & Na I Hg I2 + NaNO3 Balanced reaction is Hg (NO3)2 + 2 NaI → Hy I2 + 2NaNO3 I mole 2 mole Imole 1 mole Hg NO3)2 = 2 mole Na I

than I mole NaI I Ź mole Hg (NO3)2 0.00408 mole Nal = oroohoo mole Hg (No3l. mole Hg (NO Ē 0.00 204 mole My (NO3)2 Molarity = 0.209 Volume in L = mole 0.00204 z 0.00 976 Molan't 0:209 Tvolume in mL = 9.76 m2l & Hg (NO3)2 2 mole Nal = 1 mole My Iz Olookof mole Nal = txoroolod mole ng 12 E 0.002 ou mole Hy I2 Produced Molat mass of Hg I2 2 454. 4 glmele mass of Hg 72 produced a mole & molar mass - R S = 0.00204x45404 . - 019269 gram