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?

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:

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.

your answers are a,b,,e and f

i hope this