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

How much heat will be released when 644 g of sulfur reacts with excess O2 according to the following equation 2S3O 2SO3 H7914kJ class=

Respuesta :

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.

What is molar enthalpy of a reaction?

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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