Answered

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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.

Sagot :

Answer:

-292.8J/mmol

Explanations:

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

[tex]6HCl+6NaOH\rightarrow6NaCl+6H_2O[/tex]

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:

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

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

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