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The heat of combustion of gaseous ammonia is $ 81kcal/mol $ . How much heat will be evolved in the reaction of $ 34grams $ of ammonia with an excess amount of oxygen?
A. $ 40.5kcal $
B. $ 60.3kg $
C. $ 75.8kcal $
D. $ 81kcal $
E. $ 162kcal $

Answer
VerifiedVerified
467.4k+ views
Hint: The quantity of heat liberated when a certain amount of a substance undergoes combustion, also known as the calorific value or the energy value, can be defined as the heat of combustion. Higher calorific value, also known as gross calorific value, and higher heating value are the two forms of heat produced by burning. Lower calorific value (sometimes referred to as net calorific value or heating value).

Complete answer:
Heat of combustion $ = 81kcal/mol $
Molar Mass of Ammonia $ = 17g/mol $
 $ Number{\text{ of moles = }}\dfrac{{Mass}}{{Molar{\text{ }}mass}} $
Using the above formula,
 $ 34grams $ of ammonia corresponds to $ \dfrac{{34g}}{{17g/mol}} = 2moles $
During the reaction of $ 34grams $ ( $ 2moles $ ) of ammonia with an excess amount of oxygen, heat evolved will be $ 2mol \times 81kcal/mol = 162kcal $ .
Hence, the correct option is E. $ 162kcal $ .

Note:
It should be noted that a substance's heat of combustion can be expressed in the following units: When one mole of fuel is completely burned with oxygen, energy (in joules or kilojoules) is released. When one gram or kilogram of fuel is completely burned with oxygen, energy (in joules or kilojoules) is released. When one litre of fuel is completely burned with oxygen, energy (in joules or kilojoules) is released.