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If 49.7 mL of 1.0 M NaOH react with excess HCl creating 3046.5 J of heat,...

If 49.7 mL of 1.0 M NaOH react with excess HCl creating 3046.5 J of heat, what is the ΔHneutralization in kJ/mol?

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Hint: Answer includes 2 decimal places.

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Answer #1

Mol of NaOH reacting = M(NaOH)*V(NaOH)

= 1.0 M * 0.0497 L

= 0.0497 mol

Q = -3046.5 J

= -3.0465 KJ

Negative because heat is released

ΔHneutralization = Q / number of mol

= -3.0465 KJ / 0.0497 mol

= -61.298 KJ/mol

Answer: -61.30 KJ/mol

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