How much heat will be released when 6.44 g of sulfur reacts with excess O^2 according to the following equation? 2S+3O= 2SO^3. H=-791.4kJ

Answer:
-79.14kJ
Explanation:
2 moles of S will liberate 791.4kJ of heat
2*32kg of S = -791.4kj
6.44kg of S = X
Then find X
From the molar enthalpy value of the reaction, 79.9 kJ of heat will be released when 6.44 g of sulfur reacts with excess oxygen.
The molar enthalpy of a reaction is the amount of heat released when 1 mole of a substance reacts to form products.
From the equation of the reaction and the enthalpy value of reaction,
2 moles of sulfur reacts with 3 moles of oxygen to release 791.4kJ of heat.
Molar enthalpy of reaction = 791.4/ 2 = 395.7 kJ/mol
Moles of sulfur in 6.44 g = 6.44/32 = 0.201 moles
Heat released = 0.201 × 395.7 = 79.9 kJ
Therefore, 79.9 kJ of heat will be released when 6.44 g of sulfur reacts with excess oxygen.
Learn more about molar enthalpy at: https://brainly.com/question/25758173
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