The standard enthalpy of formation of Mn2O3 is −962.3 kJ/mol. How much heat energy is liberated when 6.5 grams of manganese are oxidized by oxygen gas to Mn2O3 at standard state conditions? Answer in units of kJ

Answer 1

2Mn + (3/2)O2 ==> Mn2O3 + 962.3 kJ

So you have 962.3 kJ heat liberated by oxidizing 2*54.94 g Mn. You want to know heat liberated by oxidizing 6.5 g. That's
962.3 kJ x (6.5/2*54.94) = ?

Answer 2

I'm plugging that into my calculator and when I plug it into the answer box it keeps telling me that the answer is incorrect.