Answer to Question #325852 in Physical Chemistry for elaine

Question

How much potassium iodide (166.00 g/mol) is required to get 1.78 g of 

mercury(I) iodide precipitate (454.39 g/mol). The unbalanced chemical equation is

Answer 1

m(HgI₂) = 1.78 g;

M(HgI₂) = 454.39 g/mol;

M(KI) = 166.00 g/mol;

n(HgI₂) = m(HgI₂)/M(HgI₂) = 1.78/454.39 = 0.004 mol;

2KI + Hg²⁺ = HgI₂ + 2K⁺;

By the chemical reaction:

n(KI) = 2 * n(HgI₂) = 2 * 0.004 = 0.008 mol;

m(KI) = n(KI) * M(KI) = 0.008 * 166.00 = 1.328 g.

Answer: 1.328 g.