Final answer:
During the combustion of 500 g of glucose, assuming full reaction with sufficient oxygen, 300 grams of H2O are produced based on stoichiometry and the known molar masses of glucose and water.
Step-by-step explanation:
The question posed involves calculating the mass of H2O produced during the combustion of glucose, C6H12O6, in the presence of O2.
Firstly, we analyze the reaction stoichiometry: each mole of glucose reacts with 6 moles of oxygen to yield 6 moles of water.
The molar masses are critical for these calculations: 180.16 g/mol for glucose and 18.015 g/mol for water.
To solve this, we first find the moles of glucose in 500 g:
moles of glucose = 500 g / 180.16 g/mol = 2.777 moles
From the stoichiometry, 1 mole of glucose produces 6 moles of water, so 2.777 moles of glucose produce:
2.777 moles * 6 = 16.662 moles of H2O
Now, we convert moles of water to grams:
mass of H2O = 16.662 moles * 18.015 g/mol = 300.037 grams
Therefore, 300 grams of H2O are produced during the combustion of 500 g of glucose, assuming there is enough O2 present to react completely.