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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
Expert Answer
Answer
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
