Final answer:
Approximately 11.67 grams of O₂ reacted with hydrogen to produce 13.1 grams of water, based on the stoichiometry of the chemical equation which states that one mole of O₂ produces two moles of H₂O.
Step-by-step explanation:
The student is asking how many grams of O₂ reacted to produce 13.1 grams of water. The balanced chemical equation for the reaction between hydrogen (H₂) and oxygen (O₂) to form water (H₂O) is:
2 H₂(g) + O₂(g) → 2 H₂O(l)
According to this equation, two moles of hydrogen react with one mole of oxygen to form two moles of water. From the molecular weights (2 x 2.02 g for H₂ and 32.0 g for O₂), we can determine that one mole of O₂ (32.0 g) will produce two moles of H₂O (2 x 18.02 g = 36.04 g). Therefore, to find out how many grams of O₂ reacted to produce 13.1 grams of H₂O, we can set up a proportion, where x is the mass of O₂:
× / 32.0 g = 13.1 g / 36.04 g
Solving for x gives us the mass of O₂ that reacted:
x = (13.1 g × 32.0 g) / 36.04 g
x ≈ 11.67 g
So, approximately 11.67 grams of O₂ reacted to produce 13.1 grams of water.