Answer to Question #132572 in General Chemistry for van sung

Question

a 2.00 g 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 0.750 g of Ag(s) were produced in this process, what should the final mass of the copper wire be?

Answer 1

Cu(s) + 2Ag⁺ = 2Ag(s) + Cu²⁺

n(Ag) = m(Ag)/M(Ag)=0.750 g/107.868 g/mol = 0.00695 mol = 1/2 n(Cu²⁺)= 0.003475 mol

m(Cu) = m₀(Cu) - m(Cu²⁺) = 2 - n(Cu²⁺)* M(Cu²⁺)= 2 - 0.003475*63.546 = 2 - 0.2208 = 1.7792 g

If we assume that the silver has settled on the wires, then 1.7792 + 0.75= 2.5292 g