It you have two rooms of hydrogen, the hydrogen is in excess and oxygen is the limiting reactant.
The balanced equation is
#"2H"_2 + "O"_2 → "2H"_2"O"#
#"Moles of O"_2 = 16 color(red)(cancel(color(black)("g O"_2))) × ("32.00 g O"_2)/(1 color(red)(cancel(color(black)("mol O"_2)))) = "0.500 mol O"_2#
#"Moles of H"_2"O" = 0.500 color(red)(cancel(color(black)("mol O"_2))) × ("2 mol H"_2"O")/(1 color(red)(cancel(color(black)("mol O"_2)))) = "1.00 mol H"_2"O"#
#"Mass of H"_2"O" = 1.00 color(red)(cancel(color(black)("mol H"_2"O"))) × (18.02 "g H"_2"O")/(1 color(red)(cancel(color(black)("mol H"_2"O")))) = "18 g H"_2"O"#