Suppose that 0.650 mol of methane, CH4(g), is reacted with 0.800 mol 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?

Question

Substance & Enthalpy of Form. (kJ/mol)
C(g) 718.4
CF4(g) -679.9
CH4(g) -74.8
H(g) 217.94
HF(g) -268.61

I have figured out the equation:
CH4(g)+4F2(g) -> CF4(g)+4HF(g)

I also figured out the enthalpy of reaction:
Delta H = -1679.54kJ

I realize that I must figure out which of the reactants are limiting. How do I do that? What do I do after I find the limiting reactant?

Answer 1

Use the coefficients in the balanced equation to convert moles CH4 to a product (say CF4) and to convert moles F2 to moles CH4. The smaller number of moles produced is the correct value and the reagent producing that value is the limiting reagent. For example, for 0.650 moles CH4, converted to moles CF4 is
0.650 moles CH4 x (1 mole CF4/1 mole CH4) = 0.650 x (1/1) = 0.650 moles CF4 produced if we started with 0.650 moles CH4 and had all of the F2 we needed.
After identifying the limiting reagent, then heat released is
q = delta Hrxn x (moles of limiting reagent/4 = ??

Answer 2

Great, my answer was correct. But, why did you divide the moles of limiting reactant by 4?

Answer 3

because CH4 and 4F_2 ratio is 1:4

Answer 4

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?