CH₃CH₂CH₂COOH(l) + 5O₂(g) → 4CO₂(g) + 4H₂O(g) ∆H = -2.18 × 10³ kJ When a 5.42-g sample of butyric acid (molar mass = 88.10 g/mol) is burned, how much energy (in kJ) is released as heat?

Answer 1

The enthalpy of reaction symbolized as  ΔH or  ΔH_rxn are defined in units of kJ/mol, so we can find the mol of butyric acid from the given mass and molar mass by dividing the mass by the molar mass.

Thus, 5.42g of butyric acid x (1/88.10g/mol) or

5.42g / 88.1g/mol will give us the number of moles of butyric acid:

which gives us .061521 moles of butyric acid.

Since the enthalpy in this reaction is -2.18x10³ kJ/mol, we can find the energy released as heat by multiplying this value by the moles of butyric acid:

-2.18x10³kJ/mol x (.061521mol) = -1.34x10² kJ.

Let me know if this helps!