1.0 M of NaOH, 1.0 M of HCl. 28 J is the heat calorimeter B. Heat of Neutralizat

Question

1.0 M of NaOH, 1.0 M of HCl. 28 J is the heat calorimeter

B. Heat of Neutralization of HCI with NaOH Show your calculations Molecular equation: HCl+ NaOH Ionic equation: Net ionic equation: ? Initial temperature of calorimeter and NaOH, T Final temperature after mixing. Tr Difference in temperature, AT-Tr-T Total masss of the solution (density 1.00g/ImL) 22.8 C 29,4 C Heat gained by solution -,xmsution xAT Heat gained by calorimeter-Cakrine·ar Heat of reaction heat gained by the solution+ heat gained by the calorimeter # moles of acid used # of moles of H2O produced Experimental value of Enthalpy of neutralization Heat of reaction/moles of H,O Use enthalpies of formation from Table 1 to calculate the theoretical AH of neutralization for the reaction of NaOH reaction with HCI Calculate the percent error

Explanation / Answer

Molecular equation,

HCl + NaOH - - - > NaCl... + H2O

Ionic equation :

H+.. +... Cl-... Na+... +.. OH-.. - - - - - > Na+.. +.. Cl-.... +H2O

Net ionic equation :

H+..+....OH-----> H2O

Heat gained by solution = m*c*dt

= * 4.17 J /g°C * 100g * 6.6 °C

= 2752.2 J

Heat gained by calorimeter = C(calorimeter) * dT

= 28 J /°C * 6.6 °C

=184.8 °C

Heat of reaction = heat gained by (solution + calorimeter)

= 2752.2 J + 184.8 J

=2937 J

- from molecular equation we can see 1 mol of acid produced 1 mol of H2O

-ENTHALPY OF NEUTRALIZATION = HEAT OF REACTION / MOL OF H20

= 2937 J / 1

=2937 J

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