Asked by Gabriella

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

Answers

Answered by DrBob222
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) = ?
Answered by Gabriella
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.
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