Posted by Bill on Sunday, November 13, 2011 at 4:00pm.
0.1850*0.400 = ?moles HCl
0.1525*0.500 = x moles NaOH
I have 0.07625 moles NaOH and 0.07400 moles HCl; therefore, 0.07400 moles H2O should be formed.
56 kJ/mol x moles H2O formed = heat released.
-4.1 KJ
That's what I have.
Related Questions
CHEMISTRY - The enthalpy of neutralization for the reaction of a strong acid ...
CHEMISTRY - The enthalpy of neutralization for the reaction of a strong acid ...
chemistry - Given that ?Hf for OH- ions is -229.6 kJ/mol, calculate the enthalpy...
College Chemistry - Given that delta Hrxn for OH- ions is -229.6 kJ/mol, ...
College - Chemistry - Given that delta Hrxn for OH- ions is -22.9 kJ/mol, ...
Chemistry - The standard enthalpies of formation of ions in aqueous solutions ...
Chemistry- college - The standard enthalpies of formation of ions in aqueous ...
Chemistry II - The enthalpy of neutralizatino for the reaction of a strong acid ...
Chemistry II - The enthalpy of neutralizatino for the reaction of a strong acid ...
Chemistry - If an equal number of moles of the weak acid HOCN and the strong ...
For Further Reading