The answer is #6.7g#.
Starting from the balanced chemical equation
#C_6H_6O_3 + 6O_2 -> 6CO_2 + 3H_2O#
we can see that we have a #1:3# mole ratio between #C_6H_6O_3# and #H_2O#; that is, for every mole of #C_6H_6O_3# used in the reaction, #3# moles of #H_2O# are produced.
SInce the reaction's percent yield is #39.0%#, we can say that not all #C_6H_6O_3# was used in the reaction <=> #O_2# is the limiting reagent.
The number of #C_6H_6O_3# moles, knowing its molar mass is #126g/(mol)#, is
#n_(C_6H_6O_3) = m_(C_6H_6O_3)/(molarmass) = (40.8g)/(126g/(mol)) = 0.32# moles
Now, if all the #C_6H_6O_3# would have reacted, the number of moles and, subsequently, the mass of #H_2O# produced would have been
#n_(H_2O) = 3 * n_(C_6H_6O_3) = 3 * 0.32 = 0.96# moles, and
#m_(H_2O) = n_(H_2O) * (molarmass) = 0.96 mol es * 18g/(mol) = 17.3g#
However, we know that the actual number of moles of #H_2O# produced is
#%yield = n_(actual)/n_(theo retic) * (100%) ->#
#n_(actual) = (%yield * n_(theo retic))/(100%) = (39.0 * 0.96)/(100%) = 0.37# moles
Therefore, the mass of #H_2O# produced is
#m_(H_2O) = n_(actual) * (molarmass) = 0.37 mol es * 18g/(mol) = 6.7g#
