# Problem: 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) calculate [NH4+] and [Cu2+] in solution after this reaction is complete.

###### FREE Expert Solution

We will do the following steps to solve the problem:

Step 1: Calculate the mol of reactants

Step 2: Determine the limiting reactant

Step 3: Determine the mass of the precipitate

Step 4: calculate [NH4+] and [Cu2+] in solution

80% (92 ratings) ###### Problem Details

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) calculate [NH4+] and [Cu2+] in solution after this reaction is complete.