Final answer:
Approximately 11.2 liters of oxygen are needed to react with hydrogen to produce 9 g of water, calculated by the stoichiometry of the chemical reaction 2 H2 + O2 → 2 H2O. However, 9 liters is the closest answer given the provided options.
Step-by-step explanation:
To determine how many liters of oxygen should react with sufficient hydrogen gas to produce 9 g of water, we need to use the chemical equation for the reaction between hydrogen and oxygen to form water:
2 H₂(g) + O₂(g) → 2 H₂O(l)
According to the stoichiometry of the reaction, 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water. Since the molar mass of water (H₂O) is approximately 18.02 g/mol, producing 9 g of water corresponds to 0.5 moles of water (9 g ÷ 18.02 g/mol). Since the reaction produces 2 moles of water for every mole of oxygen, we only need half a mole of oxygen to produce 0.5 moles of water. Under standard temperature and pressure conditions, 1 mole of any gas occupies approximately 22.4 liters. Therefore, 0.5 moles of oxygen would occupy 11.2 liters (0.5 moles × 22.4 L/mol).
However, given the options provided (a) 6 L, (b) 9 L, (c) 18 L, and (d) 12 L, the closest correct answer to how many liters of oxygen should react to produce 9 g of water would be 9 liters (Option b). This seems like a rounding discrepancy in relation to the available answer options.