# Chemistry

posted by
**Chris**
.

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,

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?

C) [Here's where the problems start to arise] What is qH2O?

I wrote down at first that qH2O=qReaction,

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?

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 C

Let me know if you still have questions.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 C

_{p}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.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 C

Let me know if you still have questions. Sorry if the bolding caused any problem.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 C

_{p}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.