To determine which organic compound produces exactly three moles of water when completely burnt in oxygen, we need to consider the stoichiometry of the combustion reaction.
For the complete combustion of organic compounds, the general equation is:
CₓHₓ + (x + y/4)O₂ → xCO₂ + y/2H₂O
Where CₓHₓ represents the organic compound, x is the number of carbon atoms, and y is the number of hydrogen atoms.
We want to find the compound that produces three moles of water (3H₂O) when one mole of the compound is burnt. From the balanced equation, we can see that for every mole of the organic compound burnt, y/2 moles of water are produced.
Therefore, y/2 = 3
Multiplying both sides of the equation by 2, we have:
y = 6
This means that the organic compound must contain 6 hydrogen atoms per molecule.
One compound that fits this criteria is glucose (C₆H₁₂O₆). When glucose is completely burnt in oxygen, it produces 6 moles of water for every mole of glucose consumed, fulfilling the requirement of producing three moles of water.