Chemistry

posted by .

Magnesium sulfate is often used it first aid hot-packs giving off heat when dissolved in water. When 2.00 g of MgSO4 dissolves in 15.0 mL of water (d= 1.00g/mL) at 25.0ºC, 1.51 kJ of heat is evolved.
A) Write the balanced equation for this reaction.
I have: MgSO4 + H2O --> MgO + H2SO4
Any problems so far?
B) Is the process exothermic?
I said yes.
C) [Here's where the problems start to arise] What is qH2O?
I wrote down at first that qH2O=qReaction, so qH2O= -1.51 kJ. But then I got to thinking, and was afraid that because MgSO4 was added to the solution, the H2O couldn't possibly account for all the heat given off in the reaction. Is my reasoning here wrong? If not, how would I find qH2O?
D) What is the final temperature of the solution? (Specific heat of water is 4.18 J/gºC)
What I had before I started thinking was:
q= m*Cp*ΔT
-1.51 kJ = 17.0g * 4.18J/gºC * ΔT
However, because I'm trying to find the final temperature of the solution, would I need to use the Cp of the solution, or is this fine?

Magnesium sulfate is often used in first aid hot-packs giving off heat when dissolved in water. When 2.00 g of MgSO4 dissolves in 15.0 mL of water (d= 1.00g/mL) at 25.0¨¬C, 1.51 kJ of heat is evolved.
A) Write the balanced equation for this reaction.
I have: MgSO4 + H2O --> MgO + H2SO4
Any problems so far?
I would think the following is more likely.
MgSO4+ 2H2O ==> Mg(OH)2 + H2SO4 but I'm not crazy about that one either. It's the lattice energy of the crystal vs the heat of hydration of the ions plus the second ionization constant of H2SO4.


B) Is the process exothermic?
I said yes. ok

C) [Here's where the problems start to arise] What is qH2O?
I wrote down at first that qH2O=qReaction,EXACTLY so qH2O= -1.51 kJ except its +1.51. But then I got to thinking, and was afraid that because MgSO4 was added to the solution, the H2O couldn't possibly account for all the heat given off in the reaction. Is my reasoning here wrong? If not, how would I find qH2O?
q for the reaction is 1.51 kJ for 2 grams MgSO4. It's the REACTION that is giving off the heat and water is absorbing the heat which makes it warm and that is what you want for a first aid pack.

D) What is the final temperature of the solution? (Specific heat of water is 4.18 J/g¨¬C)
What I had before I started thinking was:
q= m*Cp*¥ÄT
-1.51 kJ = 17.0g * 4.18J/g¨¬C * ¥ÄT
However, because I'm trying to find the final temperature of the solution, would I need to use the Cp of the solution, or is this fine?
You are on the right track.
Water ABSORBED the heat; therefore, q is +1.51 kJ. Change that to J (1510 J) and plug into q = mcdelta T.
m =15 (not 17) since it is 15.0 mL water with a density of 1.00 g/mL.And in most of these problems we assume the Cp is the same for pure water and for the solution.
Let me know if you still have questions.


I forgot and didn't turn off the bold when I should. I will try to redo that part of my answer.

C) [Here's where the problems start to arise] What is qH2O?
I wrote down at first that qH2O=qReaction,EXACTLY so qH2O= -1.51 kJ except its +1.51. But then I got to thinking, and was afraid that because MgSO4 was added to the solution, the H2O couldn't possibly account for all the heat given off in the reaction. Is my reasoning here wrong? If not, how would I find qH2O?
q for the reaction is 1.51 kJ for 2 grams MgSO4. It's the REACTION that is giving off the heat and water is absorbing the heat which makes it warm and that is what you want for a first aid pack.


D) What is the final temperature of the solution? (Specific heat of water is 4.18 J/g¨¬C)
What I had before I started thinking was:
q= m*Cp*¥ÄT
-1.51 kJ = 17.0g * 4.18J/g¨¬C * ¥ÄT
However, because I'm trying to find the final temperature of the solution, would I need to use the Cp of the solution, or is this fine?
You are on the right track.
Water ABSORBED the heat; therefore, q is +1.51 kJ. Change that to J (1510 J) and plug into q = mcdelta T.
m =15 (not 17) since it is 15.0 mL water with a density of 1.00 g/mL.And in most of these problems we assume the Cp is the same for pure water and for the solution.
Let me know if you still have questions. Sorry if the bolding caused any problem.

  • Chemistry -

    Just a question, why would the mass remain at 15?

  • Chemistry -

    Cual es el calor especifico de MgSO4?

  • Chemistry -

    need help

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chemistry

    a 5.00g sample of Al pellets (specific heat=0.89J/oC*g) and a 10.00g sample of Fe pellets (specific heat=0.45J/oC*g) are heated to 100.0oC. The mixture of hot iron Al is then dropped into 92.1g of water at 21.8oC. Calculate final temp …
  2. Chemistry

    How much energy (heat) is required to convert 248 g of water from 0oC to 154oC?
  3. Chemistry

    LiI(s) → Li+(aq) + I‾ (aq) Solid LiI dissolves as shown in the above equation. A 8.00 g sample of LiI was dissolved in 57.5 g of water. The initial temperature of the water was 21.40°C. After the compound dissolved, the …
  4. Chemistry

    This is a lab I have, but have no clue what to do. Please do out the steps so i can understand. Reactions and Data: NaOH(s)--> Na1+(aq)+ OH1-(aq) (Heat of solution of NaOH) This reaction involves adding solid NaOH to water and watching …
  5. Physcial Science

    Hey There, I have 4 questions that I am totally stumped on... Can anyone help?
  6. Chemistry HEAT OF FUSION

    What is the final temperature, in oC, after a 15.0 g piece of ice, at 0oC, is placed in a styrofoam cup with 128 g of water initially at 74.0oC. Assume there is no transfer of heat to or from the surroundings. The specific heat of …
  7. Chemistry I

    Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin plastic divider. When the divider is broken, the ammonium nitrate dissolves according to the following endothermic …
  8. chemistry

    An electric range burner weighing 614 grams is turned off after reaching a temperature of 499.0oC, and is allowed to cool down to 22.2oC. Calculate the specific heat of the burner if all the heat evolved from the burner is used to …
  9. chemistry

    KCl(s) → K+(aq) + Cl‾ (aq) Solid KCl dissolves as shown in the above equation. A 6.60 g sample of KCl was dissolved in 42.3 g of water. The initial temperature of the water was 21.90°C. After the compound dissolved, the …
  10. chem

    You mix 5.32 g of solid magnesium sulfate with 105.24 g of water inside a calorimeter. As the magnesium sulfate dissolves, the temperature of the solution rises from 24.5°C to 30.1°C. Calculate the amount of energy in kJ that was …

More Similar Questions