Question

Show Sub/Sup What are the concentrations of Cut, NH, and Cu(NH3)2 at equilibrium when 18.8 g of Cu(NO3)2 is added to 1.0 L of a 0.800 M solution of aqueous ammonia? Assume that there is no volume change upon the addition of the solid, and that the reaction goes to completion and forms Cu(NH3) K 2.0 x 1012 [Cu]- Мx [NH]= 0.4 М [Cu(NH,)0.1 M / (1 of 1) Show Approach Show Hint Submit Nei Incorrect

Answer 1

Answer! Cutz + 4NH3 C u(NH3 + 2 (aq) (aq) lag). moles of cut? (or) Cu (Noz - Mass of cu(NO3)2 Molar mass = 18.8 187.56 g/mol = 0. 100 moles. So it can give o. 100 moles of product. moles of NH₃ = Mxv. = 0.800m xllit = 0.800 moles a moles of NH3 - Con give lamole product 0.800 moles & NH₂ - Con give ? product

it can give = 0.800 a = 0.200 Moles product. So cut acts of limiting reactont and modes of Cu (NH3 + 2 = 0.100 moles [Cu (NH3272] = 0.100 moles let =o.loom

cuth +UNH₂ W Cu(NH₃)4t2 l 0.1oom -X o 0.4oom tx tx As 0.400 moles involved in Reaction only o 4oo Moles left 5 0.100-X * 0.4ootx os Kf is large (= kassociated) is less so meglect x ke - Cu (MHz) u + 2 [cat] [NH₂4 2.0x10² - 0.100 [cut?] [N.Hz]" 2.0 x 10² = 0.100 [cut2] (0.400] [cat] = 1.95 810lm, (NH3 ] = 0.400m, [Cu(NH3)4 h = o-loom.