You didn't give the reaction, but from the products given, I believe the reaction is the decomposition of hydrogen perioxide #("H"_2"O"_2")#.
Balanced equation
#"2H"_2"O"_2"##rarr##"2H"_2"O" + "O"_2"#
To determine moles #"H"_2"O"# that can be produced from #"10.0 mol O"_2"#, multiply moles #"O"_2"# by the mole ratio between #"O"_2"# and #"H"_2"O"# from the balanced equation, with moles #"H"_2"O"# in the numerator.
#10.0color(red)cancel(color(black)("mol O"_2))xx(2"mol H"_2"O")/(1color(red)cancel(color(black)("mol O"_2)))="20.0 mol H"_2"O"#
To determine the mass #"H"_2"O"# that can be produced, multiply moles #"H"_2"O"# by its molar mass #("18.016 g/mol")#.
#20.0color(red)cancel(color(black)("mol H"_2"O"))xx(18.016"g H"_2"O")/(1color(red)cancel(color(black)("mol H"_2"O")))="360. g H"_2"O"=3.60xx"10"^2" g H"_2"O"# (rounded to three significant figures)
The mass of water that can be formed is #"360. g"#, which can be written as #3.60xx10^2# #"g"#.
