Mass Of Water Produced When 1 kg Of Fat In Camel's Hump Is Metabolized
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Mass Of Water Produced When 1 kg Of Fat In Camel's Hump Is Metabolized

Tristearin - Oxygen Metabolism

The fat stored in the hump of a camel is a source of energy and water.
Calculate the mass of water produced when 1 kg of fat is metabolized. Assume the fat consists entirely of the animal fat, tristearin (C57H110O6), and that tristearin reacts with oxygen to form only carbon dioxide and water.

The strings: S7P1A15 (containership-mass).
S7P6A66 (grouping/interaction- chemical)

The math:
Pj Problem of Interest is of type containership (mass).

Tristearin- Oxygen Reaction

Balanced equation of tristearin and oxygen reaction:

2C57H110O6 + 163O2 -------> 114CO2 + 110H2O

Molar mass of C57H110O6 = 57x12 + 1x110 + 16x6 = 684 + 110 + 96 = 890 g
So, 1 kg of C57H110O6 is the weight of 1000/890 = 1.124 moles of C57H110O6.

2 moles of C57H110O6 produced 110 moles of water
So, 1.124 moles will produce (1.124 x 110)/2 = 61.82 moles of water
So, mass of water produced = 61.82 x 18 = 1112.76 g = 1.1 kg.

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