Question

If the enthalpy change for the reaction below is ΔH = 198 kJ, how many grams of sulfur dioxide are produced when 387 kJ is absorbed?

2 SO₃ (g) + 198 kJ --> 2 SO₂ (g) + O₂ (g)

SO₂ molar mass = 64.07 g/mol

SO₃ molar mass = 80.07 g/mol

Report your answer in grams with the correct number of significant figures

Answer 1

2 SO₃ (g) + 198 kJ --> 2 SO₂ (g) + O₂ (g)

SO₂ molar mass = 64.07 g/mol

SO₃ molar mass = 80.07 g/mol

According to the problem

In producing two moles of SO2 , 198 kJ is absorbed

Thus,

198 KJ/ 2 mole SO2 = 99 KJ / moles

Now amount of energy = 387 KJ

Then calculate the number of moles as follows:

Amount of energy / energy per moles

= 387 KJ/ 99 KJ / moles SO2

= 3.91 Moles SO2

Amount in g = number of moles * molar mass

= 3.91 Moles SO2 *64.07 g/mol

= 250.5 g SO2