The reaction produces 24 g of water.
Step 1. Start with the balanced equation.
#"2LiOH" → "Li"_2"O" + "H"_2"O"#
Step 2. Calculate the moles of #"LiOH"#.
#"Moles of LiOH" = 63 color(red)(cancel(color(black)("g LiOH"))) × ("1 mol LiOH")/(23.95 color(red)(cancel(color(black)("g LiOH")))) = "2.63 mol LiOH"#
Step 3. Calculate the moles of #"H"_2"O"#
#"Moles of H"_2"O" = 2.63 color(red)(cancel(color(black)("mol LiOH"))) × ("1 mol H"_2"O")/(2 color(red)(cancel(color(black)("mol LiOH")))) = "1.32 mol H"_2"O"#
4. Calculate the mass of #"H"_2"O"#.
#"Mass of H"_2"O" = 1.32 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")))) = "24 g H"_2"O"#
