Given the thermochemical equation SO2(g) + ½ O2(g) ---> SO3 (g) DH = -99.1 kJ

Looks ok to me.

To calculate the enthalpy change (ΔH) when a certain amount of substance is converted, you need to use the given thermochemical equation and the stoichiometry of the reaction. Here's how you can solve it step by step:

Step 1: Find the molar mass of SO2.
The molar mass of SO2 is calculated by adding the atomic masses of sulfur (32.06 g/mol) and oxygen (16.00 g/mol x 2) together, which totals 64.06 g/mol.

Step 2: Convert the given mass of SO2 (89.6 g) to moles.
Using the molar mass of SO2, divide the given mass by the molar mass:
89.6 g / 64.06 g/mol = 1.399 mol (rounded to three decimal places)

Step 3: Determine the moles of O2 reacting.
From the balanced equation, you can see that 1 mole of SO2 reacts with 1/2 mole of O2. Therefore, the moles of O2 involved will be half the moles of SO2:
1.399 mol / 2 = 0.6995 mol (rounded to four decimal places)

Step 4: Calculate the enthalpy change (ΔH).
Use the stoichiometric ratio between SO2 and ΔH from the thermochemical equation:
ΔH = (-99.1 kJ) * (0.6995 mol)

ΔH ≈ -69.3 kJ

So, the correct answer is 5. -69.3 kJ.

It seems you made a calculation mistake, resulting in the wrong answer of -139 kJ. Double-check your calculations to confirm.