Fred wanted to reclaim silver from metal waste (iron, copper, nickle, and silver) reclaimed from teh floor of a jewelry shop. He dissolved 159.7 g of the metal waste in nitric acid and then diluted it with water. he then added 10 mL concentrated HCL and precipitated all the silver as the chloride. Upon filtration and drying, he had 42.729g of precipitate. calculate the amount of pure silver in w/w% of the original sample.
2006-10-05 12:19:18 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
atomic mass of Ag / (atomic mass of Ag + atomic mass of Cl) =
weight percentage of Ag in AgCl
x 42.729g = how much Ag
/159.7g = w/w % of Ag in original
2006-10-05 12:32:11 · answer #1 · answered by feanor 7 · 0⤊ 0⤋
this is just a (two) simple percentage problems:
Percent (%) = (part / whole) x 100%
= ( 42.729 g AgCl / 159.7 g gunk) x 100%
= 0.2676 x 100 %
= 26.76 % AgCl
[stop with 4 significant figures]
Now calculate the percent of Ag in AgCl,
%Ag = (at wt of Ag / formula wt of AgCl) x 100%
Sorry, don't have my periodic table with me. But that's how you do it.
2006-10-05 12:33:22 · answer #2 · answered by MrZ 6 · 0⤊ 0⤋