Question

Aqueous solutions of copper (II) bromide and silver (1) acetate react to form solid

silver (1) bromide and aqueous copper (II) acetate according to the UNBALANCED
reaction below.
CuBr2 (aq) + AGCH3CO2 (aq)
-
AgBr (s) + Cu(CH3CO2)2 (aq)
How many grams of silver (1) bromide will form if 30.0 mL of 0.499 M copper (II)
bromide react with excess silver (1) acetate?

Answer 1

5.63 g

Step-by-step explanation:

Step 1: Write the balanced equation

CuBr₂(aq) + 2 AgCH₃CO₂(aq) ⇒ 2 AgBr(s) + Cu(CH₃CO₂)₂(aq)

Step 2: Calculate the reacting moles of copper (II) bromide

30.0 mL of 0.499 M CuBr₂ react. The reacting moles of CuBr₂ are:

`0.0300L * (0.499mol)/(L) = 0.0150mol`

Step 3: Calculate the moles formed of silver (I) bromide

The molar ratio of CuBr₂ to AgBr is 1:2. The moles formed of AgBr are 2/1 × 0.0150 mol = 0.0300 mol.

Step 4: Calculate the mass corresponding to 0.0300 mol of AgBr

The molar mass of AgBr is 187.77 g/mol.

`0.0300 mol * (187.77g)/(mol) = 5.63 g`