The standard enthalpy of formation of Mn2O3 is −962.3 kJ/mol. How much heat energy is liberated when 6.4 grams of man- ganese are oxidized by oxygen gas to Mn2O3 at standard state conditions?

Question

Answer in units of kJ

Answer 1

We first write the balanced chemical reaction:
4 Mn + 3 O2 -> 2 Mn2O3 ; ΔH = -962.3 kJ/mol Mn2O3
We then get the number of moles Mn by dividing the given mass by molar mass of Mn. The molar mass of Mn is 54.94 g/mol (get a periodic table and search for it).
6.4 / 54.94 = 0.1165 mol Mn
From the reaction, for every 4 moles of Mn, 2 moles of Mn2O3 are produced. Thus,
0.1165 mol Mn * (2 mol Mn2O3 / 4 mol Mn) = 0.05825 mol Mn2O3
Finally, we multiply this to ΔH:
0.05825 mol Mn2O3 * -962.3 kJ/mol Mn2O3 = -56.05 kJ

Hope this helps :3