Question

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

Answer 1

Answer:

5.63 g

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:

$0.0300L \times \frac{0.499mol}{L} = 0.0150mol$

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.

$0.0300 mol \times \frac{187.77g}{mol} = 5.63 g$