2SO2(g) + O2(g) ⟶ 2SO3(g), △H = -198 kJ What is the enthalpy change that occurs when 58.0 g SO2 reacts with excess oxygen in the above reaction?

Enthalpy change = -98.6kJ

Explanation:

Based on the reaction:

2SO₂(g) + O₂(g) ⟶ 2SO₃(g), △H = -198 kJ

When 2 moles of SO₂ (Molar mass: 64.066g/mol) reacts with 1 mole of oxygen, enthalpy change is -198kJ

To know enthalpy change of 58.0g of SO₂ with excess oxygen we need to calculate moles of SO₂ thus:

58.0g SO₂ ₓ (1mol / 64.066g) = 0.905 moles SO₂

If 2 moles of SO₂ produce an enthalpy change of -198kJ, 0.905 moles SO₂ will produce:

0.905 moles SO₂ ₓ (-198kJ / 2 moles SO₂) =

Enthalpy change = -98.6kJ

i think it is d. e because the next one is igneous.

i dont know

explanation: