Question

1.250g of an alloy which is 90.00% Ag and 10.00% Cu is dissolved in nitric acid yielding AgNO3 and Cu(NO3)2 solution. The nitrates are converted to Ag2SO4 and CuSO4 by heating with H2SO4. To the solution of the sulfates is added a piece of copper wire which reacts with the Ag2SO4 to precipitate all the silver. The reaction is: Ag2SO4 (aq) + Cu (s) -----> 2 Ag (s) + CuSO4 (aq). The resulting solution is treated with excess zinc to precipitate all of the copper (II) ion in the solution. What weight of copper will be precipitated?

Answer 1

Step-by-step explanation:

Amount of silver

= 1.25 x .9 = 1.125

amount of copper = 1.25 x .1 = .125

equivalent weight of silver = 108

gram equivalent of silver in the alloy = 1.125 / 108 = .01041

This will precipitate in the form of copper in equivalent amount of copper

equivalent of copper precipitated = .01041

grm of copper for silver = .01041 x equivalent weight of copper

= .01041 x 63.5 / 2 = .3305 gram

Total copper precipitated

= .125 + .3305

= .4555 gm