1. Consider the following: HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (liq) ∆H = -57.62 kJ/mol If a 25.0 mL of 0.144 M HCl (aq) at 25oC is added to 20.0 mL of 0.132 M NaOH (aq) at 25oC, calculate the final temperature of the contents. Assume the volumes are additive and that the resulting salt water solution has a density of 1.04 g/mL with a specific heat capacity of 3.93 J/goC. Note that you will also need to determine the limiting reactant.

25.82°C

Explanation:

Based on the reaction, 1 mole of HCl and 1 mole of NaOH reacts, that means the reaction is 1:1. The moles of each compound are:

Moles HCl:

0.025L * (0.144mol/L) = 0.0036 moles HCl

Moles NaOH:

0.020L * (0.132mol/L) = 0.00264 moles NaOH

Thus, moles of reaction are 0.00264 moles

The heat released in a calorimeter is obtained using the equation:

Q = m*c*ΔT

Where Q is heat released in the reaction:

0.00264 moles * (-57.62kJ/mol) = 0.1521kJ = 152.1J of reaction

m is mass of the solution:

25.0mL + 20.0mL = 45mL * (1.04g/mL) = 46.8g

c is specific heat of the solution:

3.93J/gºC

And ΔT is change in temperature.

Solving for ΔT:

Q /mc = ΔT

151.2J / 46.8g*3.93J°C = 0.82°C = ΔT = Final temperature - Initial temperature.

Final temperature = 0.82°C + 25°C =

idk

explanation: