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

Question

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

How much heat is released when 4.00 mol iron is reacted with excess O2?

How much heat is released when 1.00 mol Fe2O3 is produced?

How much heat is released when 1.00 iron is reacted with excess O2?

How much heat is released when 10.0g Fe and 2.00g O2 are reacted?

I've been looking at this for a long time, and I'm really stuck on starting each question...

Answer 1

Then you don't understand the question because that equation tells you everything you ever wanted to know and then some.
Here is what it says. It tells you that if 4 mols Fe reacts with 3 mols O2 it will produce 2 mols Fe2O3 and 1652 kJ of heat.

So the first question is 4 mols Fe and xs O2, how much heat? What is your answer to that?

Answer 2

This is just basic stoichiometry.

From the reaction, this is the basic information that you will need:

4 moles of Fe = -1652 kJ
3 moles of O2 = -1652 kJ
and
2 moles of Fe2O3= -1652 kJ

So, for your first question:

4 moles of Fe = -1652 kJ

For your second question:

1.00 mol Fe2O3*(-1652 kJ/2 moles of Fe2O3)=???
****Answer should contain three significant figures.

You can answer the third question.

For your first question, you have to determine the limiting reagent:

10.0g Fe*(1 mole/55.845g)= moles of Fe

2.00g O2*(1 mole/32.00g)= moles of O2

Eyeballing it, I think O2 is the limiting reagent.

Proceed as you would for the second problem.

Answer 3

For your ***Last question, you have to determine the limiting reagent:

10.0g Fe*(1 mole/55.845g)= moles of Fe

2.00g O2*(1 mole/32.00g)= moles of O2

Eyeballing it, I think O2 is the limiting reagent.

Proceed as you would for the second problem.