The amount of water that will be produced is 50.36 grams
Stoichiometric problems
The metabolism of glucose is represented by the following equation:
[tex]C_6H_1_2O_6(s)+6O_2(g)--- > 6CO_2(g)+6H_2O(g)[/tex]
The mole ratio of glucose metabolized to the water produced is 1:6.
Mole of 84.0 g glucose = 84/180.156 = 0.4662 moles
Equivalent mole of water = 0.4662 x 6 = 2.7975 moles
Mass of 2.7975 moles water = 2.7975 x 18 = 50.36 grams
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