Ice Calorimetry: Determine DeltaH for Ca + HCl Rxn

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Discussion Overview

The discussion revolves around calculating the enthalpy change (DeltaH) for the reaction between metallic calcium (Ca) and hydrochloric acid (HCl) using an ice calorimeter. The focus is on the methodology and calculations involved in determining DeltaH based on the heat absorbed from melting ice during the reaction.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant presents a calculation for DeltaH based on the amount of ice melted and the heat of fusion, concluding with a value of 521 kJ/mol for the reaction.
  • Another participant expresses agreement with the calculations presented.
  • A separate participant requests additional information about ice calorimeters, indicating a need for further understanding of the experimental setup.

Areas of Agreement / Disagreement

There is a partial agreement on the calculations presented, but the discussion does not reach a consensus on the correctness of the final DeltaH value or the methodology used.

Contextual Notes

The calculations depend on the assumptions regarding the heat of fusion and the stoichiometry of the reaction. There may be unresolved steps in the calculation process that could affect the final result.

Who May Find This Useful

Students or individuals interested in calorimetry, thermochemistry, and reaction enthalpy calculations may find this discussion relevant.

Mitchtwitchita
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Is there anybody that can tell me if I'm doing this question right?

Metallic Ca reacts with hydrochloric acid according to the following equation:
Ca(s) + 2 HCl(aq) ---> 2 CaCl2 (aq) + H2(g)
The enthalpy change associated with this reaction can be determined with an ice calorimeter. If 13.59 g of ice are melted when 0.3487 g of Ca reacts with excess hydrochloric acid, what is deltaH (in kJ/mol Ca) for the reaction written above? The heat of fusion of ice is 333 J/g.

qrxn + qfusion = 0
qrxn = ndeltaHrxn
=(13.59 g H2O)(0.333 kJ/g)
=4.53 kJ

0.3487 g Ca X 1 mol Ca/40.08 g Ca = 8.70 x 10^-3 mol

4.53 KJ/8.70 x 10^-3 mol = 521 kJ/mol?

Does this look right?
 
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Looks good to me.
 
Thanks Chemisttree!
 
thack you

please if anyone who know any thing about ice calorimeter please go to this subject in chemstry fourm ((i need information about ice calorimeter please help me.))

thanx much
 

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