Consider the following reaction.

2 HCl(aq) + Ba(OH)2(aq) BaCl2(aq) + 2 H2O(l) H = -118 kJ

Calculate the heat when 112.4 mL of 0.500 M HCl is mixed with 300.0 mL of 0.515 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25.0°C and that the final mixture has a mass of 412.4 g and a specific heat capacity of 4.18 J/°C·g, calculate the final temperature of the mixture.

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To calculate the final temperature of the mixture, we need to use the principle of conservation of energy.

First, let's calculate the heat released during the reaction. We know that the reaction releases 118 kJ of heat when one mole of HCl reacts with one mole of Ba(OH)2.

Step 1: Calculate the moles of reactants used:
Moles of HCl = (volume in liters) x (molarity)
= (112.4 mL / 1000 mL/L) x 0.500 mol/L
= 0.0562 mol

Moles of Ba(OH)2 = (volume in liters) x (molarity)
= (300.0 mL / 1000 mL/L) x 0.515 mol/L
= 0.1545 mol

Step 2: Determine the limiting reactant:
Since the stoichiometric ratio of HCl to Ba(OH)2 is 2:1, and we have less Ba(OH)2 moles than HCl moles, Ba(OH)2 is the limiting reactant.

Step 3: Calculate the heat released:
Heat released = moles of limiting reactant x heat of reaction
= 0.1545 mol x (-118 kJ/mol)
= -18.219 kJ

Step 4: Convert the heat released to joules:
Heat released in joules = -18.219 kJ x (1000 J/1 kJ)
= -18219 J

Now, let's calculate the temperature change of the solution.

Step 5: Calculate the heat capacity of the solution:
Heat capacity = (mass of solution) x (specific heat capacity)
= 412.4 g x 4.18 J/°C·g
= 1720.632 J/°C

Step 6: Calculate the temperature change using the equation:
Heat released = Heat capacity x temperature change
Therefore, Temperature change = Heat released / Heat capacity
Temperature change = -18219 J / 1720.632 J/°C
= -10.59 °C

Step 7: Calculate the final temperature:
Final temperature = Initial temperature + Temperature change
= 25.0 °C - 10.59 °C
≈ 14.41 °C

Therefore, the final temperature of the mixture is approximately 14.41 °C.