Final answer:
To find the mass of water produced when 6.85 g of O₂ reacts with excess H₂, use the balanced chemical equation and stoichiometry. The mass of water produced is 7.71 g.
Step-by-step explanation:
To find the mass of water produced when 6.85 g of O₂ reacts with excess H₂, we need to use the balanced chemical equation and stoichiometry.
The balanced equation is: 2H₂(g) + O₂(g) → 2H₂O(l)
From the equation, we can see that 1 mole of O₂ produces 2 moles of H₂O. Using the molar mass of O₂ (32.0 g/mol), we can calculate the number of moles of O₂: 6.85 g / 32.0 g/mol = 0.214 mol
Since the mole ratio is 1:2 for O₂:H₂O, the number of moles of H₂O produced will be twice the number of moles of O₂: 2 * 0.214 mol = 0.428 mol. Finally, we can calculate the mass of H₂O produced using the molar mass of H₂O (18.02 g/mol): 0.428 mol * 18.02 g/mol = 7.71 g.