Assuming all the reactants react, the limiting reactant in this case is H2, since the amount of O2 present is greater than the amount needed to react with all the H2. The balanced chemical equation for the reaction is:
2H2 + O2 → 2H2O
Using stoichiometry, we can determine that the maximum amount of water that can be produced is 10.0g H2 x (1 mol H2 / 2.016g H2) x (2 mol H2O / 2 mol H2) x (18.015g H2O / 1 mol H2O) = 89.1g H2O.
The excess reactant in this case is O2. To determine the mass of O2 that remains after the reaction is complete, we can use stoichiometry again. Since we know that all the H2 has reacted, we can use the amount of H2O produced to determine the amount of O2 that reacted. The amount of O2 that reacted is 5.0g O2 x (1 mol O2 / 32.00g O2) x (2 mol H2O / 1 mol O2) x (18.015g H2O / 1 mol H2O) = 5.63g O2. Therefore, the mass of O2 that remains after the reaction is complete is 5.0g O2 - 5.63g O2 = -0.63g O2. However, this result is not physically possible, since it implies that we started with less O2 than we actually did. Therefore, we can conclude that all the O2 has reacted and none remains after the reaction is complete.