Question

Aqueous solutions of copper (II) bromide and silver (I) acetate react to form solid silver (I) 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 (I) bromide will form if 40.0 mL of 0.155 M copper (II) bromide react with excess silver (I) acetate?

Answer 1

The balance reaction is:

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

moles of CuBr2 reacting = M*V

= 0.155 M * 0.040 L

= 0.0062 mol

from balanced reaction above,

moles of AgBr formed = 2*moles of CuBr2 reacting

= 2*0.0062 mol

= 0.0124 mol

Molar mass of AgBr = 1*MM(Ag) + 1*MM(Br)

= 1*107.9 + 1*79.9

= 187.8 g/mol

we have below equation to be used:

mass of AgBr,

m = number of mol * molar mass

= 1.24*10^-2 mol * 187.8 g/mol

= 2.33 g

Answer: 2.33 g