The following thermochemical equation is for the reaction of iron(s) with chlorine(g) to form iron(III) chloride.

2Fe(s) + 3Cl2(g) 2FeCl3(s) H = -799 kJ
How many grams of Fe(s) would have to react to produce 107 kJ of energy?

I am not sure how to start this.

You get 799 kJ of energy with 2*55.85g Fe(2 moles Fe). So how many grams would you need to obtain just 107 kJ?

Yea I could have used the help on this question instead of answering a question with a question. I needed to know the set up, clearly I'm not entirely sure. I see you do this on a few posts Dr. Bob

To determine the number of grams of Fe(s) required to produce 107 kJ of energy, we need to use the thermochemical equation and the given enthalpy change.

First, we can calculate the energy change corresponding to the reaction of 1 mole of Fe(s) using the given enthalpy change. The coefficient of Fe(s) in the balanced equation is 2, so the energy change for 2 moles of Fe(s) would be -799 kJ.

Next, we can set up a proportion to find the number of grams of Fe(s) corresponding to 107 kJ of energy:

(2 moles Fe(s) / -799 kJ) = (x moles Fe(s) / 107 kJ)

To solve for x, we can cross-multiply and rearrange the equation:

x moles Fe(s) = (2 moles Fe(s) / -799 kJ) * 107 kJ
x moles Fe(s) = (2 moles Fe(s) * 107 kJ) / -799 kJ

Now, we can convert moles of Fe(s) to grams using the molar mass of iron (Fe), which is approximately 55.845 g/mol:

mass Fe(s) = x moles Fe(s) * molar mass Fe(s)
mass Fe(s) = (x moles Fe(s)) * (55.845 g/mol)

Finally, we can substitute x in the equation to get the final answer in grams:

mass Fe(s) = [(2 moles Fe(s) * 107 kJ) / -799 kJ] * (55.845 g/mol)

By solving the equation, you will find the number of grams of Fe(s) required to produce 107 kJ of energy.