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Answer to Question #129954 in General Chemistry for Mary

Question #129954
Hi,
A neutralization experiment using NaOH and HCl produced 986 J. The limiting reactant was NaOH, which was present in 1.94×10-2 moles.
Calculate the enthalpy change of neutralization (in kJ/mol).

Please Give your answer to three significant figures. Thanks
1
Expert's answer
2020-08-28T06:31:06-0400

Enthalpy change of neutralization=heatchange/molesofNaOH =Q÷n = (986J÷ 1.94×10^-2moles)÷1000 =-50.8kJ/mol


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