Question

When 65.00 ml of 0.182 M CuBr2 is mixed with 55.00 ml of 0.147 M (NH4)2CrO4, the following reaction takes place: (NH4)2CrO4 (aq) + CuBr2 (aq) --> CuCrO4 (s) + 2 NH4Br (aq) Part A). What is the limiting reactant? Show work. Part B) determine the mass in grams of precipitate that could theoretically be formed. Show work. Part c) list all spectator ions Part D) calculate [NH4+] and [Cu2+] in solution after this reaction is complete.

Answer 1

Ans. Moles of CuBr2 = Molarity x Volume (in Liters)

                                    = 0.182 M x 0.065 L                                                    ; [1L = 1000 mL]

                                    = (0.182 mol/ L) x 0.065 L                                           ; [1 M = 1 mol/ L]

                                    = 0.01183 mol

Moles of (NH4)2CrO4 = Molarity x Volume (in Liters)

                                    = 0.147 M x 0.055 L                                       

                                    = 0.008085 mol

Experimental Molar ratio of reactants = CuBr2 : (NH4)2CrO4 = 0.01183 mol : 0.008085 mol

Given:

Balanced reaction: (NH4)2CrO4(aq) + CuBr2(aq) ------> CuCrO4(s) + 2 NH4Br(aq)

Stoichiometry: 1 mol of (NH4)2CrO4 reacts with 1 mol of CuBr2 to produce 1 mol CuCrO4 and e mol NH4Br.

Theoretical molar ratio of reactants = CuBr2 : (NH4)2CrO4 = 1: 1

Now,

Convert experimental ratio into a form similar to theoretical ratio so at least one digit is common in both the ratio.

For that, divide the experimental ration by factor 0.01183-

So, CuBr2 : (NH4)2CrO4 = (0.01183 mol : 0.008085 mol) / 0.01183 = 1 : 0.6834

Thus, normalized experimental ratio = 1 : 0.6834

Compare the theoretical and normalized ratio, the number of moles of (NH4)2CrO4 is lesser than 1.0 while that of CuBr2 remains constant.

Thus, : (NH4)2CrO4 is the limiting reagent.

Ans. B. 1 mol : (NH4)2CrO4 produces 1 mol CuCrO4(s) – the precipitate.

So, moles of precipitate (CuCro4) = moles of limiting reagent, (NH4)2CrO4

                                                = 0.008085 mol

Mass of precipitate formed = Moles of CuCrO4 precipitate x Molar mass

                                                = 0.008085 mol x (179.5397 g/ mol)

                                                = 1.45 g         

Ans. C. Spectator ions = NH₄⁺ , Br^-

An ion that remains the same on either side of the reaction is called a spectator ion. Note that these ions remain in the same form. While Cu²⁺, and CrO₄^2- change their form as they form solid precipitate from soluble ions.

Ans. D. Balanced reaction: (NH4)2CrO4(aq) + CuBr2(aq) ------> CuCrO4(s) + 2 NH4Br(aq)

Stoichiometry: All the products form according to 0.008085 mol of the limiting reactant (NH4)2CrO4(aq).

D1. Moles of NH₄⁺ ion = (2 x moles of NH4Br) + Moles of (NH4)2CrO4 remaining unreacted

Or, Moles of NH₄⁺ ion = 2 x 0.008085 mol     ; [all (NH4)2CrO4 used up- limiting reagent]

Hence, moles of NH₄⁺ ion = 0.01617 mole

Now, [NH₄⁺] = moles of NH₄⁺/ total volume of solution in liters

                        0.01617 mole / (0.120 L)

                        = 0.13475 M

Note: Total volume of solution = 65.0 mL (CuBr2) + 55.0 mL of (NH4)2CrO4

                                    = 120.0 mL = 0.120 L

D2. Balanced reaction: (NH4)2CrO4(aq) + CuBr2(aq) ------> CuCrO4(s) + 2 NH4Br(aq)

Stoichiometry: All the products form according to 0.008085 mol of the limiting reactant (NH4)2CrO4(aq).

Moles of Cu²⁺ = Initial moles of CuBr2 – Moles of CuCrO4

                        = 0.01183 mol - 0.008085 mol    

                        = 0.003745 mol

Now, [Cu²⁺] = 0.003745 mol / 0.120 L = 0.0312 M

Note: Cu²⁺ is only present in form of reaming CuBr2(aq). The Cu-atoms in form of solid CuCrO4 are NOT ionized.