sample calculations

# sample calculations - = 3022.64 J Moles of water produced...

This preview shows pages 1–2. Sign up to view the full content.

A. Heat of Neutralization for HCl and NaOH for Trial 1 T for reaction, o C = (T f -T 1 ) = 31.0 o C – 24.0 o C = 7 o C Heat gained by solution (J) = ( T x 100g x 4.18J/g o C) = 7 o C x 100g x 4.18J/g o C = 2926.0 J Heat absorbed by calorimeter (J) = ( T x Calorimeter Heat Capacity) = 7 o C x 13.52 J/ o C = 96.64 J Total heat released by reaction (J) = (heat gained by solution, J) + (Heat absorbed by calorimeter, J) = 2926.0 J + 96.64 J

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: = 3022.64 J Moles of water produced (moles) by using the HCl information = 0.05 L x . 1 0 moles of HClL x 1 mol of H2O1 mol of HCl =0.05 mol of H 2 O Molar heat of reaction, H rxn in kJ = (total heat released / moles of water produced) = . . 3022 64J0 05moles = 60452.8 J = 60.45 kJ Average value of H rxn = . + . + . 60 45 kJ 65 59 kJ 65 50 kJ3 = 63.88 kJ...
View Full Document

## This note was uploaded on 03/18/2010 for the course COS 101121501 taught by Professor Langner during the Fall '08 term at RIT.

### Page1 / 2

sample calculations - = 3022.64 J Moles of water produced...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online