Consider the reaction \(2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H=-118 \mathrm{~kJ}\) Calculate the heat when \(100.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{HCl}\) is mixed with \(300.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\). Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of \(400.0 \mathrm{~g}\) and a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\), calculate the final temperature of the mixture.

Let's find the initial moles of both reactants: 1. Moles of HCl = Volume (L) × Molarity = 0.100 L × 0.500 M = 0.050 mol 2. Moles of Ba(OH)₂ = Volume (L) × Molarity = 0.300 L × 0.100 M = 0.030 mol Now, let's check the reaction's stoichiometry to find out which reactant will run out first. The reaction is given as: \[2 \mathrm{HCl} + \mathrm{Ba(OH)}_{2} \longrightarrow \mathrm{BaCl}_{2} + 2 \mathrm{H}_{2} \mathrm{O}\] This indicates that 2 moles of HCl react with 1 mole of Ba(OH)₂. So, we will divide the moles of each reactant by their respective stoichiometric coefficients and find the limiting reactant: 1. Moles of HCl divided by its coefficient = 0.050 mol / 2 = 0.025 2. Moles of Ba(OH)₂ divided by its coefficient = 0.030 mol / 1 = 0.030 Since 0.025 < 0.030, HCl is the limiting reactant.

As HCl is the limiting reactant, we must use its moles in the reaction. Moles of HCl involved in the reaction are 0.050 moles.

To calculate the heat generated by the reaction, we must use the enthalpy change of the reaction, which is -118 kJ. Since this enthalpy change corresponds to the given reaction's stoichiometry, we must first find the moles of HCl that correspond to the enthalpy change: \(\frac{\Delta H}{2 \, \text{moles }\mathrm{HCl}} = \frac{-118 \,\text{kJ}}{2} = -59 \,\mathrm{kJ/mol}\) Now, we use the moles of HCl involved in the reaction to find the total heat generated: Heat generated by the reaction = moles of HCl × -59 kJ/mol = 0.050 mol × -59 kJ/mol = -2.95 kJ = -2950 J

We are given that the mass of the final mixture is 400.0 g and its specific heat capacity is 4.18 J/($$^{\circ}\)C·g). Using the heat generated by the reaction (-2950 J), we can calculate the change in temperature of the mixture using the following formula: \[\Delta T=\frac{-q}{mc}\] Where q is the heat generated, m is the mass of the mixture, and c is its specific heat capacity: \[\Delta T=\frac{-(-2950 \, \text{J})}{400.0 \, \text{g} \times 4.18 \,\mathrm{J}/^{\circ}\mathrm{C} \cdot \mathrm{g}} = \frac{2950}{1672}= 1.765 \,\,^{\circ}\mathrm{C}\]

Lastly, to find the final temperature of the mixture, we will add the change in temperature to the initial temperature, which is 25.0$$^{\circ}$$C: Final temperature = Initial temperature + Change in temperature = 25.0$$^{\circ}$$C + 1.765$$^{\circ}$$C = 26.765$$^{\circ}$$C Thus, the final temperature of the mixture is approximately 26.8$$^{\circ}$$C.