**Balanced equation**

#"BaO(s) + H"_2"SO"_4("aq")"##rarr##"BaSO"_4("s") + "H"_2"O(l)"#

The assumption is that #"H"_2"SO"_4"# is in excess. We can determine the moles #"BaO"# required by multiplying moles #"H"_2"O"# produced by the mole ratio between #"BaO"# and #"H"_2"O"# from the balanced equation with moles #"BaO"# in the numerator. This will cancel moles #"H"_2"O"# and leave moles #"BaO"#.

#2.9color(red)cancel(color(black)("mol H"_2"O"))xx(1"mol BaO")/(1color(red)cancel(color(black)("mol H"_2"O")))="2.9 mol BaO"#

To determine the mass of #"2.9 mol BaO"#, multiply by its molar mass #("153.326 g/mol")#.

#2.9color(red)cancel(color(black)("mol BaO"))xx(153.326"g BaO")/(1color(red)cancel(color(black)("mol BaO")))="440 g BaO"# rounded to two significant figures