Enthalpy of Reaction Calculation | Magnesium and Water | Redox Reaction

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SUMMARY

The discussion focuses on calculating the enthalpy of reaction for magnesium reacting with water in an ice calorimeter setup. The balanced chemical equation is Mg(s) + 2H2O(l) → Mg(OH)2 + H2(g), confirming it as a redox reaction where magnesium is oxidized and hydrogen is reduced. The enthalpy calculation involves determining the mass of ice melted, which is found to be 2.099g, and using the enthalpy of fusion for water (6.01 kJ/mol) to compute the heat of the reaction. The final step requires dividing the total heat by the moles of magnesium to obtain the enthalpy change.

PREREQUISITES
  • Understanding of redox reactions and their identification.
  • Familiarity with the concept of enthalpy and calorimetry.
  • Knowledge of the enthalpy of fusion for water (6.01 kJ/mol).
  • Ability to perform stoichiometric calculations involving moles and mass.
NEXT STEPS
  • Calculate the enthalpy change for the reaction using the derived mass of water and the enthalpy of fusion.
  • Explore the concept of calorimetry and its applications in thermochemistry.
  • Learn about the principles of redox reactions and their significance in chemical processes.
  • Investigate the properties of magnesium and its reactions with water in different conditions.
USEFUL FOR

Chemistry students, educators, and anyone interested in thermochemistry and reaction energetics will benefit from this discussion.

hunter457
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Homework Statement



In an ice calorimeter, the reaction between 0.0272g of magnesium with excess water was carried out. The volume of water in the calorimeter decreased by 0.19mL during the reaction.
(a) Write the equation for the reaction which occurs. Is it a redox reaction?
(b) Calculate the enthalpy of the reaction (in KJ/mol).


Homework Equations


q=mc∆t


The Attempt at a Solution


(a) I figured out that the equation is Mg(s) + 2H2O(l) Mg(OH)2 + H2(g) and that it is a redox reaction (hydrogen is reduced, magnesium is oxidized).

(b) ∆V=V2-V1=Volume of water produced by melting-volume of ice melted

mass of ice melted=mass of water produced= m

At 0degrees celsius, density of water=1g/mL, density of ice=0.917g/mL

∆V=(mass of water produced/density of water) - (mass of ice melter/density of ice)
= m(1/1 - 1/0.917)
= m x (-0.0905)
m= ∆V/-0.0905= -0.19/(-0.0905)= 2.099g of ice melted

∆Hsystem=0
∆Hsystem=∆HvesselA+∆HvesselB=0
∆HvesselA = -∆HvesselB

***Is this right so far? Where do I go from here?***
 
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I think you need to find the enthalpy for melting ice into water: 6.01 kJ. Then take the mass you calculated, divide by the molar mass of water to get moles of water, and then multiply the moles by 6.01 kJ to get the heat of water (dH = q/n). Then take the q and divide by the moles of magnesium to find the heat of reaction. I think that's how you do it.
 

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