Question

# *A solution contains Ag+ and Hg2+ ions. The addition of 0.100 L of 1.06 M NaI...

*A solution contains Ag+ and Hg2+ ions. The addition of 0.100 L of 1.06 M NaI solution is just enough to precipitate all the ions as AgI and HgI2. The total mass of the precipitate is 24.6 g .

*Find the mass of AgI in the precipitate.

Express your answer to two significant figures and include the appropriate units

number of moles of I- added = M(NaI) * V (NaI)
= 1.06 M * 0.1 L
= 0.106 mol

Let number of moles of AgI be n1 and HgI2 be n2

number of moles of I in product = n1 +2*n2
use:
n1+2n2 = 0.106 mol
n2 = (0.106-n1)/2

mass of AgI = Molar mass of AgI * number of moles of AgI
mass of AgI = 234.77 * n1

mass of HgI2 = Molar mass of HgI2 * number of moles of HgI2
mass of AgI = 454.4 * n2

Total mass of precipitate = 24.6 g
234.77 * n1 + 454.4 * n2 = 24.6
234.77 * n1 + 454.4 * (0.106-n1)/2 = 24.6
469.54* n1 + 48.1664 -454.4*n1 = 49.2
15.14* n1= 1.0336
n1 = 0.0683 mol

Molar mass of AgI = 234.77 g/mol

Mass of AgI = number of moles * molar mass
= 0.0683 * 234.77
= 16 g