Question

Suppose that 0.410 mol of methane, CH4(g), is reacted with 0.560 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

Answer 1

Answer:

The balanced equation for this reaction will be

                            $CH4 + 4F2$    →  $CF4 + 4HF$

We can see that 1 mole of methane requires 4 moles of fluorine but we have 0.41 moles of CH4 and 0.56mole of F2

So using the unitary method we will get that

• 1 mole of CH4 → 4 mole of 4 mole of fluorine

• 0.41 mole of methane  →  4*0.41 = 1.64 mole of fluorine for complete reaction

but we have only 0.56 mole of fluorine that means fluorine is the limiting reagent and the product will only be formed by only this amount of fluorine.

• 4 moles of  fluorine →  1 mole of CF4

• 0.56 mole →  $\frac{1}{4} * 0.56$ = 0.14mole of CF4

• 4 moles of fluorine →  4 moles of HF

• 0.56 mole of fluorine →  0.56 mole of HF

now to find the heat released we have the formula as

DELTA H = n * Delta H of product - n *delta H of reactant

where n is the moles of the reactant and product.

note: since no information is given about the enthalpies of the species we leave it on general equation also you need to add the product side enthalpy of the species present and similarly on the product side.