50.0 mL each of 1.0 M HCl and 1.0 M NaOH, at room temperature (20.0 OC) are mixed. The temperature of the resulting NaCl solution increases to 27.5 OC. The density of the resulting NaCl solution is 1.02 g/mL. The specific heat of the resulting NaCl solution is 4.06 J/g OC Calculate the Heat of Neutralization of HCl(aq) and NaOH(aq) in KJ/mol NaCl produced

-62.12kJ/mol is heat of neutralization

Explanation:

The neutralization reaction of HCl and NaOH is:

HCl + NaOH → NaCl + H₂O + HEAT

An acid that reacts with a base producing a salt and water

You can find the released heat of the reaction  -heat of neutralization- (Released heat per mole of reaction) using the formula:

Q = C×m×ΔT

Where Q is heat, C specific heat of the solution (4.06J/gºC), m its mass of the solution and ΔT change in temperature (27.5ºC-20.0ºC = 7.5ºC).

The mass of the solution can be found with the volume of the solution (50.0mL of HCl solution + 50.0mL of NaOH solution = 100.0mL) and its density (1.02g/mL), as follows:

100.0mL × (1.02g / mL) = 102g of solution.

Replacing, heat produced in the reaction was:

Q = C×m×ΔT

Q = 4.06J/gºC×102g×7.5ºC

Q = 3106J = 3.106kJ of heat are released.

There are 50.0mL ×1M = 50.0mmoles = 0.0500 moles of HCl and NaOH that reacts releasing 3.106kJ of heat. That means heat of neutralization is:

3.106kJ / 0.0500mol of reaction =

-62.12kJ/mol is heat of neutralization

The - is because heat is released, absorbed heat has a + sign

Aconductor easily transfer heat

Standing in the sunlight and feeling the heat on your face (uv radiation is entering the skin, converting to infrared radiation, and is absorbed by your tissues)orstanding near a fire and feeling the heat on your skin (infrared radiation transfer)