Carol asks…what mass of NaCl is needed to precipitate the silver ions?What mass of NaCl is needed to precipitate the silver ions from 18.0 mL of 0.180 M AgNO3 solution?Nancy answers:First write a balanced chemical equation:
NaCl + AgNO3 ------> AgCl(s) + NaNO3
Next, figure out the number of moles of AgNO3. We can do this with the formula M = mmols / mL. Since you know M and mL, plug those values in and solve for mmol:
0.180 = mmols / 18mL = 3.24 mmol of AgNO3
Looking at the balanced
