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.
Just a question, why would the mass remain at 15?
Cual es el calor especifico de MgSO4?
need help
Where did you get 17g ?
A) The balanced equation you have written is not correct. The correct balanced equation for the reaction is: MgSO4 + 7H2O → Mg(OH)2 + H2SO4.
B) Yes, the process is exothermic.
C) qH2O in this case refers to the amount of heat absorbed by the water. Since the water is the solvent in which the magnesium sulfate dissolves, it accounts for the majority of the heat given off in the reaction. So, qH2O=qReaction, and in this case qH2O is equal to +1.51 kJ.
D) To find the final temperature of the solution, you can use the equation q = mCΔT, where q is the heat absorbed by the water (1.51 kJ converted to J), m is the mass of the water (15.0 g), C is the specific heat of water (4.18 J/g°C), and ΔT is the change in temperature. Rearranging the equation to solve for ΔT, you can then find the final temperature of the solution.
A) The balanced equation for the reaction between magnesium sulfate (MgSO4) and water (H2O) is: MgSO4 + 2H2O → Mg(OH)2 + H2SO4.
B) Yes, the process is exothermic because heat is evolved during the dissolution of magnesium sulfate in water.
C) The value of qH2O is the heat absorbed by the water during the reaction. In this case, the heat evolved by the reaction is given as 1.51 kJ. Since the reaction is exothermic, the water absorbs this heat. Therefore, qH2O is +1.51 kJ, not -1.51 kJ.
D) To find the final temperature of the solution, you can use the equation q = m * Cp * ∆T, where q is the heat exchanged, m is the mass of the water, Cp is the specific heat of water, and ∆T is the change in temperature. In this case, the heat exchanged (q) is 1510 J (converted from 1.51 kJ), the mass of the water (m) is 15.0 g (based on the volume and density given), and the specific heat of water (Cp) is 4.18 J/g°C. Since you are trying to find the final temperature (∆T), you can rearrange the equation to solve for ∆T: ∆T = q / (m * Cp). Plugging in the values, we get ∆T = 1510 J / (15.0 g * 4.18 J/g°C), which gives the value of ∆T. Adding ∆T to the initial temperature of 25.0°C will give you the final temperature of the solution.