In a coffee-cup calorimeter, \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) and \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) are mixed to yield the following reaction: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) $$ The two solutions were initially at \(22.60^{\circ} \mathrm{C}\), and the final temperature is \(23.40^{\circ} \mathrm{C}\). Calculate the heat that accompanies this reaction in \(\mathrm{kJ} / \mathrm{mol}\) of \(\mathrm{AgCl}\) formed. Assume that the combined solution has a mass of \(100.0 \mathrm{~g}\) and a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g} .\)

We are given that 50.0 mL of 0.100 M \(\mathrm{AgNO}_{3}\) solution is mixed with 50.0 mL of 0.100 M \(\mathrm{HCl}\) solution. The reaction is: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) $$ Since the reaction has a 1:1 stoichiometry, the amounts of \(\mathrm{Ag}^{+}\) and \(\mathrm{Cl}^{-}\) that react are equal. So we can find the moles of \(\mathrm{AgCl}\) formed by finding the moles of either \(\mathrm{Ag}^{+}\) or \(\mathrm{Cl}^{-}\) in the initial solutions and taking the lowest amount as the limiting reagent. Moles of \(\mathrm{Ag}^{+}\) = Molarity × Volume = 0.100 mol/L × 0.0500 L = 0.00500 mol Moles of \(\mathrm{Cl}^{-}\) = Molarity × Volume = 0.100 mol/L × 0.0500 L = 0.00500 mol Both have equal moles, so either can be considered the limiting reagent. Thus, 0.00500 mol of \(\mathrm{AgCl}\) is formed.

The two solutions are initially at 22.60 \(^\circ\mathrm{C}\) and finally reach a temperature of 23.40 \(^\circ\mathrm{C}\). The mass of the combined solution is 100.0 g, and its specific heat capacity is 4.18 \(\mathrm{~J} / { }^{\circ} \mathrm{C} \cdot \mathrm{g}\). To calculate the heat transferred, we will use the formula: q = mcΔT where: - q is the heat transferred - m is the mass of the combined solution (100.0 g) - c is the specific heat capacity (4.18 \(\mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\)) - ΔT is the change in temperature (23.40 \(^\circ\mathrm{C}\) - 22.60 \(^\circ\mathrm{C}\)) q = (100.0 g) × (4.18 \(\mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\)) × (23.40 \(^\circ\mathrm{C}\) - 22.60 \(^\circ\mathrm{C}\)) q = (100.0 g) × (4.18 \(\mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\)) × (0.80 \(^\circ\mathrm{C}\)) q = 334.4 J

Now we have the total heat transferred (q) and the moles of \(\mathrm{AgCl}\) formed. To calculate the heat per mole of \(\mathrm{AgCl}\) formed, we will use the formula: Heat per mole of AgCl = \(\frac{q}{n}\) where: - q is the heat transferred (334.4 J) - n is the moles of \(\mathrm{AgCl}\) formed (0.00500 mol) Heat per mole of AgCl = \(\frac{334.4 \mathrm{~J}}{0.00500 \mathrm{~mol}}\) = 66880 J/mol Now we will convert the heat per mole of AgCl from J/mol to kJ/mol: 66880 J/mol × \(\frac{1 \mathrm{~kJ}}{1000 \mathrm{~J}}\) = 66.9 \(\mathrm{kJ} / \mathrm{mol}\) The heat that accompanies this reaction is 66.9 \(\mathrm{kJ} / \mathrm{mol}\) of \(\mathrm{AgCl}\) formed.