Suppose that 0.650mol of methane, CH4 (g), is reacted with 0.800mol of fluorine, F2 (g), forming CF4 (g) and HF (g) as sole products. Assuming that the reaction occurs at constant pressure, how much heat is released?

Answer 1

This is a combination limiting reagent problem as well as a heat production problem. Do it in steps. First the limiting reagent and the amount of product formed.
Write and balance the equation.
CH4 + 4F2 ==> CF4 + 4HF

1. Convert 0.650 mol CH4 to mols CF4. Use the coefficients in the balanced equation to do it.
0.650 mol CH4 x (1 mol CF4/1 mol CH4) = 0.650 mols CF4 produced.

Convert 0.800 mol F2 to mols CF4.
0.800 mol F2 x (1 mol CF4/4 mol F2) = 0.800 x 1/4 = 0.200

In limiting reagent problem you are likely to obtain values for the product that are different (as is this case) so one of them must be wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Thus F2 is the limiting reagent at 0200 mol.

Next we must determine how much of the CH4 reacted and how much HF is produced. That is done with coefficients, too.
0.800 mol F2 x (1 mol CH4/4 mol F2) = 0.800 x 1/4 = 0.200 mol CH4 used.
We know 0.200 mol CF4 is produced; therefore, we must have produced 4 x 0.200 = 0.800 mol HF.

Now we plug data into the delta H rxn using delta Ho formation values that you probably can find in your text.
dHrxn = (mols*dH products) - (mols *dH reactants). dHrxn is the heat produced by that reaction.