The standard enthalpy of formation of mn2o3 is â962.3 kj/mol. how much heat energy is liberated when 7.9 grams of manganese are oxidized by oxygen gas to mn2o3 at standard state conditions? answer in units of kj.

Answer : -69.286 KJ heat energy is liberated.

Solution : Given,

Standard enthalpy of formation : When one mole of a substance is formed from its constituent elements.

  (for 1 mole)

Mass of Mn = 7.9 g

Molar mass of Mn = 54.93 g/mole

First we have to calculate the moles of Mn.

The formation reaction of is represented as,

For this reaction, the standard enthalpy of formation is,

From the reaction, we conclude that

4 moles of Mn releases heat energy = -1924.6 KJ

0.144 moles of Mn releases heat energy =

Therefore, -69.286 KJ heat energy is liberated.

the answer your looking for is c).