Question

A 2.000g piece of copper wire was used to precipitate silver ions as silver from a solution. When the copper metal reacts, Cu2+ ions are formed, which are soluble in water. If .750g of Ag were produced in this process, what should the final mass of the copper wire be?

Answer 1

m = 2 g of copper wire

2Ag+ + Cu(s) --> Ag(s) + Cu+2

if

m = 0.75 g of Ag are produced...

find mass of copper wire

mol of Ag+ reacted = mass/MW = .075/107.86820 = 0.006952929 mol of Ag

now...

2 mol of Ag = 1 mol of Cu

0.006952929 mol --> 1/2*0.006952929 = 0.003476 mol of Cu

then

mass of Cu lost = mol *MW = 0.003476*63.5460 = 0.2209 g

mass of Ag gained = 0.75

Total mass change = 0.75 - 0.2209 = 0.5291 g

Total mass = Mas sinitial + Change in mass = 2+0.5291 = 2.5291 g