The standard enthalpy of formation of
Mn2O3 is −962.3 kJ/mol. How much heat
energy is liberated when 4.7 grams of manganese are oxidized by oxygen gas to Mn2O3
at standard state conditions?
Answer in units of kJ
To calculate the heat energy liberated when 4.7 grams of manganese (Mn) are oxidized by oxygen gas (O2) to form Mn2O3, we need to use the concept of molar mass and stoichiometry.
1. Determine the molar mass of manganese (Mn):
The molar mass of Mn is 54.94 g/mol.
2. Convert the given mass of manganese (4.7 grams) to moles:
Moles = Mass / Molar mass
Moles = 4.7 g / 54.94 g/mol
Moles ≈ 0.0853 mol
3. Determine the stoichiometry of the reaction:
The balanced chemical equation for the oxidation of manganese by oxygen gas is:
4 Mn + 3 O2 -> 2 Mn2O3
From the balanced equation, we can see that:
4 moles of Mn react with 3 moles of O2 to produce 2 moles of Mn2O3.
4. Calculate the moles of Mn2O3 produced:
Using stoichiometry, we can calculate the moles of Mn2O3 produced:
Moles of Mn2O3 = (Moles of Mn / 4) * (2/1)
Moles of Mn2O3 = 0.0853 mol / 4 * 2/1
Moles of Mn2O3 ≈ 0.0427 mol
5. Calculate the heat energy liberated:
The standard enthalpy of formation of Mn2O3 is given as -962.3 kJ/mol. This means that for each mole of Mn2O3 formed, -962.3 kJ of heat energy is liberated.
Heat energy liberated = Moles of Mn2O3 * Standard enthalpy of formation
Heat energy liberated = 0.0427 mol * (-962.3 kJ/mol)
Heat energy liberated ≈ -41.1 kJ
Since the heat energy liberated is a negative value, we can express it as approximately 41.1 kJ (liberated).