Question

Using the ΔH -48.8 kJ/mmol, calculate the enthalpy change if you neutralized six moles of HCl with an equal number of moles NaOH to one decimal place.

Answer 1

Answer:

-292.8J/mmol

Explanations:

Given the reaction of NaCl and NaOH expressed by the chemical equation:

$6HCl+6NaOH\rightarrow6NaCl+6H_2O$

Amount of heat released when one mole of strong acid reacts with one mole of the strong base is -48.8 kJ/mmol.

If 6 moles of the sample acid reacts with 6 moles of NaOH, the required enthalpy cheange is expressed as:

$\begin{gathered} Amount\text{ of heat released }=6\times-48.8 \\ Amount\text{ of heat released}=-292.8kJ\text{/mmol} \end{gathered}$

Hence the enthalpy change if you neutralized six moles of HCl with an equal number of moles NaOH is -292.8J/mmol