What mass of silver chloride can be prepared by the reaction of 140.0 mL of 0.19 M silver nitrate with
140.0 mL of 0.25 M calcium chloride?
Mass = 3.81 g
The equation for this precipitation reaction using the formulas of the reactants and products is
$$2 AgNO_3 (aq) + CaCl_2 (aq) \rightarrow 2AgCl(s) + Ca(NO_3)_2 (aq)$$
Solid AgCl forms according to the following net ionic equation:
$$Ag^+ (aq) + Cl^- (aq) \rightarrow AgCl(s)$$
The number of moles of these ions in initial solutions are:
$$140.0 mL \times \frac{1 L}{1000 mL} \times \frac{0.19 mol}{L} = 0.0266 mol Ag^+$$
$$140.0 mL \times \frac{1 L}{1000 mL} \times 2 \times \frac{0.25 mol}{L} = 0.0700 mol Cl^-$$
Because $$Ag^+$$ and $$Cl^-$$ react in a 1:1 ratio, the amount of $$Ag^+$$ will be limiting (0.0266 mole $$Ag^+$$ is less than 0.0700 mole of $$Cl^-$$. Since $$Ag^+$$ is limiting, only 0.0266 mole of solid AgCl will be formed.
Therefore, the mass of AgCl can be calculated:
$$0.0266 mol AgCl \times \frac{143.4 g AgCl}{1 mol AgCl} = 3.81 = 3.8 g AgCl$$
Calculate the concentrations of each ion remaining in solution after precipitation is complete.
(If no ions remain, leave the box blank and click on Submit.)
Concentration of $$Ag^+$$ = M
Concentration of $$Cl^-$$ = M
Concentration of $$Ca^{2+}$$ = M
Concentration of $$NO_3^-$$ = M
