posted by Jo on .
A 12 mL sample of 0.4 M silver nitrate solution is mixed with 8 mL of 0.8 M sodium chloride solution. Determine the identity of precipitate and the number of grams produced.
12 mL x 0.4M AgNO3 = 4.8 millimols.
8 mL x 0.8M NaCl = 6.4 mmols.
AgNO3 + NaCl => AgCl + NaNO3
This is a limiting reagent problem since amounts are given for BOTH reactants. You can tell, since the reactants and products are 1:1:1:1 that AgNO3 is the limiting reagent so you will produce 4.8 mmillimols (0.0048 mols) AgCl and convert that to grams. g = mols x molar mass.