The answer is #236g#.
Starting from the balanced chemical equation
#2C_3H_6 + 9O_2 -> 6CO_2 + 6H_2O#
we can see that we have a #2:6# (or a #1:3#) mole ratio between #C_3H_6# and #CO_2#; that is, for every mole of #C_3H_6# used in the reaction, #3# moles of #CO_2# are produced.
Now, since the reaction's percent yield is #81.3%#, we can say that not all #C_3H_6# was used in the reaction <=> #O_2# is the limiting reagent.
The number of #C_3H_6# moles, knowing its molar mass is #42g/(mol)#, is
#n_(C_3H_6) = m_(C_3H_6)/(molarmass) = (91.3g)/(42g/(mol)) = 2.2# moles.
If all the #C_3H_6# would have reacted, the number of moles and, subsequently, the mass of #CO_2# would have been
#n_(CO_2) = n_(C_3H_6) * 3 = 6.6# moles, and #m_(CO_2) = n_(CO_2) * (molarmass) = 6.6 mol es * 44g/(mol) = 290g#
However, we know that the actual number of #CO_2# moles produced is
#%yield = n_(actual)/n_(the o re tic) * 100% -> #
#n_(actual) = (%yield * n_(the o retic))/(100%) = (81.3% * 6.6)/(100%) = 5.37# moles
Therefore, the actual mass of #CO_2# produced is
#m_(CO_2) = n_(actual) * 44g/(mol) = 5.37 mol es* 44 g/(mol) = 236g#
