CH4 + 4F2 ==> CF4 + 4HF
Determine the limiting reagent (LR).
mols CH4 = 0.490
mols F = 0.640
F must be the LR.
Then
dHo rxn = (n*dHo products) = (n*dHo reactants) and that will be for the reaction as written. You didn't have 4 mols F2, you had only 0.640. Correct for that. Post your work if you get stuck AND post the dHo values you used.
Suppose that 0.490 mol of methane, CH4(g), is reacted with 0.640 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?
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