[lilmul123]

Hi, I have a couple thermochemistry questions, and I would love some help on how to do some.  Here are two examples of problems I have.  Can anyone please explain how to go about doing them?  Thanks!

The overall reaction in commercial heat packs can be represented as shown below.

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)      H = -1652 kJ

    (a) How much heat is released when 3.34 mol iron is reacted with excess O2?

and..

In a coffee-cup calorimeter, 1.60 g of NH4NO3 is mixed with 78.0 g of water at an initial temperature of 25.00°C. After dissolution of the salt, the final temperature of the calorimeter contents is 23.83°C. Assuming the solution has a heat capacity of 4.18 J/°Cg and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of NH4NO3 in units of kJ/mol.

[Borek]

Please read forum rules.

[lilmul123]

I don't see how exactly explaining how I was doing them will help.  I know for sure I was doing it wrong, but here goes nothing.

For the first question, I tried multiplying the moles of Fe by the enthalpy of the reaction.  That was wrong.  I looked in my Zumdahl book, I was pretty sure that was how it was done, but it's not.

For two, I tried to use q=mC(DT) to find q.  I thought the problem may have been with my masses.  So I tried 78 grams as a mass, 1.6 as a mass, and a combination of the two as a mass; all of them were wrong. 

Also, for both, how I determine whether the answer will be negative or not?

[Borek]

I don't see how exactly explaining how I was doing them will help.  I know for sure I was doing it wrong, but here goes nothing.

At least we know you have tried

For the first question, I tried multiplying the moles of Fe by the enthalpy of the reaction.  That was wrong.  I looked in my Zumdahl book, I was pretty sure that was how it was done, but it's not.

4 moles of iron per reaction as stated - you are using 3.34.

[lilmul123]

I'm using 3.34 mol because that's what question A is asking for. 

See, I'm totally confused and frustrated!   

[Borek]

For the first question, I tried multiplying the moles of Fe by the enthalpy of the reaction.

If you will use 4 moles of Fe, you will get -1652 kJ. If you will use 1 mole of Fe, you will get ¹/₄ of -1652 kJ. You are using 3.34 moles...

[lilmul123]

For the first question, I tried multiplying the moles of Fe by the enthalpy of the reaction.

If you will use 4 moles of Fe, you will get -1652 kJ. If you will use 1 mole of Fe, you will get ¹/₄ of -1652 kJ. You are using 3.34 moles...

that makes perfect sense!  thank you.

now...if anyone could help with the calorimetry problem, that would make me burst in tears of happiness

[Borek]

For two, I tried to use q=mC(DT) to find q.  I thought the problem may have been with my masses.  So I tried 78 grams as a mass, 1.6 as a mass, and a combination of the two as a mass; all of them were wrong.

You should use sum of both masses. Show more of what you did.