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

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be