What mass of AgCl will precipitate when .050 L of a 0.0500 M solution of AgNO3 reacts with 25.0 mL of .0330 M NaCl ? AgNO3(aq) + NaCl (aq)  NaNO3(aq) + AgCl (s)

Answer 1

AgNO3(aq) + NaCl(aq) ==> AgCl(s) + NaNO3(aq) ... balanced equation

Find the limiting reactant by dividing moles of each reactant by its own coefficient:

moles AgNO3 present = 0.050 L x 0.0500 mol/L = 0.0025 moles AgNO3 (÷1 -> 0.0025)

moles NaCl present = 0.025 L x 0.0330 mol/L = 0.000825 moles NaCl (÷1 -> 0.000825)

NaCl is limiting and will determine how much product can be formed

mass AgCl formed = 0.000825 mol NaCl x 1 mol AgCl / mol NaCl x 143.3 g/mol = 0.118 g AgCl (3 sig.figs.)