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:
[tex]0.0300L \times \frac{0.499mol}{L} = 0.0150mol[/tex]
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.
[tex]0.0300 mol \times \frac{187.77g}{mol} = 5.63 g[/tex]
