Final answer:
When 500 g of glucose combusts in the presence of 500 g of O₂, 300.01 grams of water are produced.
Step-by-step explanation:
To calculate how many grams of H₂O are produced during the combustion of 500 g of glucose (C₆H₁₂O₆) in the presence of 500 g of O₂, we need to look at the balanced equation for the combustion of glucose. From the information provided, we know that:
- One mole of glucose reacts with 6 moles of O₂ to produce 6 moles of CO₂ and 6 moles of H₂O.
- The molar mass of glucose is 180.16 g/mol, O₂ is 31.9988 g/mol, and H₂O is 18.015 g/mol.
Given the molar mass of glucose, the 500 g of glucose is equivalent to 500 g / 180.16 g/mol = 2.777 moles of glucose. The stoichiometry of the reaction is a 1:6 ratio of glucose to water. Therefore, 2.777 moles of glucose would produce 6 * 2.777 moles of water. That is 16.662 moles of water.
Now, to find the mass of water produced:
Mass of H₂O = moles of H₂O * molar mass of H₂O
Mass of H₂O = 16.662 moles * 18.015 g/mol
Mass of H₂O = 300.01 g
Thus, 300.01 grams of water are produced from the combustion of 500 g of glucose.