## MCAT Question A Day - 2/16/14 - Answer!

**The heat of combustion of gaseous ammonia, NH**

_{3}(*g*), is 81 kcal/mole. How much heat is released in the reaction of 34 grams of ammonia with excess oxygen?

**A.**40.5 kcal

**B.**60.3 kcal

**C.**75.8 kcal

**D.**81 kcal

**E.**162 kcal

*E. This question combines some basic stoichiometry with thermodynamic values for a reaction. We are given the ΔHcomb per mole of ammonia and the mass of ammonia in grams. Applying dimensional analysis, we must convert the mass of ammonia into moles, then multiply by kcal/mol to find the total number of kcal. Ammonia has a formula weight of 14 + 3(1) = 17 g/mol, therefore:*

*(34 g) / (17 g/mol) = 2 mol*

*(2 mol) * (81 kcal/mol) = 162 kcal.*

*Note that it is not necessary to come up with a balanced equation for the combustion reaction to do this problem.*

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