To determine the limiting reactant and the mass of water produced, we need to use the balanced chemical equation for the reaction between hydrogen gas and oxygen gas to form water:2 H g + O g 2 HO l First, we need to convert the mass of each reactant to moles using their molar masses:Molar mass of H = 2.02 g/molMolar mass of O = 32.00 g/molMoles of H = 25 g / 2.02 g/mol = 12.38 molMoles of O = 35 g / 32.00 g/mol = 1.09 molNow, we need to determine the mole ratio of the reactants:Mole ratio of H to O = 12.38 mol / 1.09 mol = 11.36The balanced chemical equation tells us that the mole ratio of H to O should be 2:1 for the reaction to proceed completely. Since the actual mole ratio is 11.36, which is greater than 2, this means that there is an excess of hydrogen gas, and oxygen gas is the limiting reactant.Now, we can calculate the mass of water produced using the stoichiometry of the balanced equation:Moles of HO produced = 2 * moles of O = 2 * 1.09 mol = 2.18 molMolar mass of HO = 18.02 g/molMass of HO produced = 2.18 mol * 18.02 g/mol = 39.24 gSo, the mass of water that can be produced is 39.24 grams, and the limiting reactant is oxygen gas.