Zinc is determined by precipitating and weighing as Zn2Fe(CN)6. a. What weight of zinc is contained in a sample that gives 0.348 g precipitate? b. What weight of precipitate would be formed from 0.500 g of Zinc?

Answer 1

MM = molar mass
AM = atomic mass

a.
0.348g Zn2Fe(CN)6 x [2*AM Zn/MM Zn2Fe(CN)6]

b.
0.5 g Zn x (MM Zn2Fe(CN)6/2*AM Zn)

Answer 2

(a) 0.26 g (b) 1.30 g