What mass of silver chloride can be prepared by the reaction of \(100.0 \mathrm{~mL}\) of \(0.20 \mathrm{M}\) silver nitrate with \(100.0 \mathrm{~mL}\) of \(0.15 \mathrm{M}\) calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete.

First, we need the balanced chemical equation for the reaction between silver nitrate (AgNO₃) and calcium chloride (CaCl₂) to produce silver chloride (AgCl) and calcium nitrate (Ca(NO₃)₂). The balanced chemical equation is: \(2 \mathrm{AgNO_3} + \mathrm{CaCl_2} \rightarrow 2 \mathrm{AgCl} + \mathrm{Ca(NO_3)_2}\)

We need to find out which reactant will run out first. To do this, we will use the volume and concentration of each reactant to calculate the moles of each reactant and compare their mole ratio to the stoichiometry in the balanced chemical equation. Moles of silver nitrate (AgNO₃): \(100.0 \mathrm{~mL} \times 0.20 \mathrm{M} = 0.020 \mathrm{mol}\) Moles of calcium chloride (CaCl₂): \(100.0 \mathrm{~mL} \times 0.15 \mathrm{M} = 0.015 \mathrm{mol}\) Now, divide the moles of each reactant by their coefficients in the balanced chemical equation: Silver nitrate: \(\frac{0.020 \mathrm{mol}}{2} = 0.010\) Calcium chloride: \(\frac{0.015 \mathrm{mol}}{1} = 0.015\) Since 0.010 is smaller than 0.015, silver nitrate is the limiting reactant, and it will determine the mass of silver chloride formed.

Now, we can use stoichiometry to calculate the mass of silver chloride (AgCl) formed by the complete reaction of the limiting reactant, silver nitrate (AgNO₃). Moles of silver chloride produced = moles of silver nitrate used (based on the balanced chemical equation) = \(0.020 \mathrm{mol}\) Now, we need to determine the molar mass of silver chloride (AgCl): \(M_{\mathrm{AgCl}} = M_{\mathrm{Ag}} + M_{\mathrm{Cl}} = 107.87 + 35.45 = 143.32 \mathrm{~g/mol}\) Mass of silver chloride produced = moles × molar mass of AgCl = \(0.020 \mathrm{mol} \times 143.32 \mathrm{~g/mol} = 2.866 \mathrm{~g}\)

We know the reaction uses all of the silver nitrate (AgNO₃) and produces 0.020 moles of silver chloride (AgCl). To find the moles of calcium chloride (CaCl₂) remaining after the reaction, we need to calculate the moles of calcium chloride (CaCl₂) that reacted with silver nitrate (AgNO₃): Moles of CaCl₂ reacted = 0.010 (based on the stoichiometry in the balanced chemical equation) Moles of CaCl₂ remaining in the solution = 0.015 - 0.010 = 0.005 mol Total volume of solution = 100.0 mL + 100.0 mL = 200.0 mL = 0.200 L Concentration of Cl⁻ ions in solution = \(\frac{\mathrm{Moles~of~CaCl_2~remaining}}{\mathrm{Total~volume~of~solution}} = \frac{0.005 \mathrm{~mol}}{0.200 \mathrm{~L}} = 0.025 \mathrm{M}\) Concentration of Ca²⁺ ions in solution = 0.025 M Since there are no more Ag⁺ ions in the solution (all of them reacted with Cl⁻ ions to form insoluble AgCl), the concentration of Ag⁺ ions in solution is 0 M. Concentration of NO₃⁻ ions in solution: AgNO₃ initially contributed 0.020 mol of NO₃⁻ ions. Since all the AgNO₃ reacted, these ions remain in the solution. Concentration of NO₃⁻ ions in solution = \(\frac{0.020 \mathrm{~mol}}{0.200 \mathrm{~L}} = 0.10 \mathrm{M}\) Final concentrations of each ion in the solution: - Ag⁺: 0 M - Cl⁻: 0.025 M - Ca²⁺: 0.025 M - NO₃⁻: 0.10 M