What is the theoretical yield (in g) and percent yield of water in the reaction (%) ?

This is the balanced equation:

#C_2H_5OH + 3O_2 -> 2CO_2 + 3H_2O#

From this, we know that #1# mole of #C_2H_5OH# will react with #3# moles of #O_2# to form #3# moles of #H_2O#.

The question tells us that there were #"4.64 mL"# of #C_2H_5OH# and that the density was #"0.789 g/mL"#. So, the number of grams of #C_2H_5OH# that reacted must have been:

#"4.64 mL" xx "0.789 g/mL" = "3.66 g"#

To find the number of moles of #C_2H_5OH#, we just need to divide this number by the molar mass of #C_2H_5OH#:

#"3.66 g"/(C xx 2 + H xx 5 + O + H) = "3.66 g"/"46.07 g/mol" = "0.0794 mol"#

Similarly, to find the number of moles of #O_2# that reacted, we'll need to divide #"15.75 g"# by the molar mass of #O_2#:

#"15.75 g"/(O xx 2) = "3.70 g"/"32.00 g/mol" = "0.492 mol"#

To react fully with #"0.492 mol"# of #O_2#, #"0.492 mol" -: 3 = "0.164 mol"# of #C_2H_5OH# needs to react.
We only have #"0.0794 mol"#, so #C_2H_5OH# would be the limiting reactant in this case.

Since #C_2H_5OH# is the limiting reactant, we'll be using #"0.0794 mol"# to calculate the number of #H_2O# produced—for every #1# mole of #C_2H_5OH# that reacts, #3# moles of #H_2O# will be produced.

So, the number of moles of #H_2O# will be #"0.0794 mol" xx 3 = "0.238 mol"#.

Multiplying this by the mass of #1# mole of #H_2O# will give us the mass of #"0.238"# moles:

#"0.238 mol" xx (Hxx2 + O) = "0.238 mol" xx "18.02 g/mol" = "4.29 g"#

That's our theoretical yield—we expect #"4.29 g"# of #H_2O# to be produced. But, after running the experiment, only #"3.70 mL" xx "1.00 g/mL" = "3.70 g"# was collected.
So, the percent yield would be:

#"actual yield"/"theoretical yield" = "3.70 g"/"4.29 g" = 0.863 = 86.3%#