Where Did the 400g in Enthalpy Change of Neutralization Come From?

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Discussion Overview

The discussion revolves around the calculation of the enthalpy change of neutralization, specifically questioning the origin of the 400g mass used in the calculations. Participants explore the definitions and calculations related to enthalpy changes in chemical reactions, focusing on the context of neutralization reactions between acids and bases.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the source of the 400g mass in the enthalpy change calculation, noting their own calculations yield a total of 144g from the reactants and products involved.
  • Another participant asks for clarification on whether the heat was measured for the water produced or the entire solution.
  • A follow-up inquiry seeks to determine the volume of the solution and its density to better understand the mass calculation.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the 400g figure, with no consensus reached on its origin or the assumptions needed for the calculations.

Contextual Notes

Participants mention the need for approximations regarding solution density and volume, indicating that these factors could influence the mass calculations but do not resolve the discrepancies in the figures presented.

Who May Find This Useful

This discussion may be useful for students and educators in chemistry, particularly those interested in enthalpy changes, neutralization reactions, and the calculations involved in thermochemical processes.

Iwanttolearnphysics
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44
Reaction score
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Homework Statement
Calculate the enthalpy change of neutralization for the following reaction
Relevant Equations
q = mcΔT
Hi, everyone! There's a question I found on website I'm using and the answer key here is given. My question is this, where did the 400g come from?

According to the definition of enthalpy of neutralization (chem libretexts), the standard enthalpy change of neutralization is the enthalpy change when solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water.

I highlighted 1 mole of water because that's what I used to solve the problem.

This is what I did step by step:
  • First, I wrote the balanced chemical equation. I know that 0.800 mol of NaOH and 0.800 mol of HCl was consumed. From the balanced chemical equation, I concluded that 0.800 mol of water must have been produced too.
  • 0.800 mol of water is equivalent to 14.4g of water.
  • I used q = mcΔT and I wrote:
    • q = 14.4g x 4.18 x (27.7-25.1)
    • q = 156.50 J of heat was released per 0.800 mol
    • If I want to get per mol, then 156.50J/0.800 mol = 195.62 J/mol

What did I do wrong? Thank you.

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What was heated up: water produced, or whole solution?
 
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Likes   Reactions: Iwanttolearnphysics and DrJohn
Sorry for the late reply. I was trying to find ways to solve this problem on my own and I still don't get it. Yeah, it was the whole solution that was heated up. But I still don't get why it's 400g?

I used the molar masses to calculate the masses in grams of the following compounds and these are what I got:
  • 1 mol of NaOH = 32g
  • 1 mol of HCl = 36g
  • 1 mol of H2O = 18g
  • 1 mol of NaCl = 58g

32 + 36 + 18 + 58 = 144g

Where did the 400g come from?
 
What is the volume of the solution?

What is the solution density (yes, we need some approximations here)?

What is the solution mass?
 

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