2HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2H2O(l) ΔH = –118 kJ Calculate the heat when 250.0 mL of 0.500 M HCl is mixed 500.0 mL of 0.500 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25.0 oC and that the final mixture has mass of 750.0 g and a specific heat capacity of 4.18 J oC–1g–1, calculate the final temperature (in oC) of the mixture.

Heat = 7375J

Final temperature of the mixture = 27.35°C

Explanation:

In the reaction:

2HCl(aq) + Ba(OH)₂(aq) → BaCl₂(aq) + 2H₂O(l) ΔH = –118 kJ

When 2 moles of HCl reacts with excess of Ba(OH)₂ there are released 118kJ.

In the reaction, moles of HCl and Ba(OH)₂ that reacts are:

Moles HCl = 0.250L ₓ (0.500 moles / L) = 0.125 moles HCl

Moles Ba(OH)₂ = 0.500L ₓ (0.500 moles / L) = 0.250 moles Ba(OH)₂

For a complete reaction of 0.125 moles of HCl you need:

0.125 mol HCl ₓ (1 mole Ba(OH)₂ / 2 moles HCl) = 0.0625 moles Ba(OH)₂

As you have 0.250 moles of Ba(OH)₂, this reactant is in excess

2 moles of HCl that react release 118kJ, 0.125 moles of HCl release:

0.125 moles HCl ₓ (118kJ / 2 moles) = 7.375kJ =

The heat released can be obtained with the formula:

Q = C×m×ΔT

Where Q is heat, C specific heat of the solution, m its mass and ΔT change in temperature.

Replacing:

Q = C×m×ΔT

7375J = 4.18J/g°C×750.0g×ΔT

2.35°C = ΔT

As ΔT = Final T - Initial T:

2.35°C = Final T - 25.0°C

i believe the answer is three water molecules