We can start by writing the chemical equation for this reaction:
#2"LiOH(aq)" + "CO"_2"(g)" rarr "Li"_2"CO"_3 "(aq)" + "H"_2"O(l)"#
A simple way to find the limiting reagent is to calculate the number of moles of #"Li"_2"CO"_3# that can form from each #4.00"mol LiOH"# and #3.20"mol CO"_2#. Whichever reagent produces the lesser amount of #"Li"_2"CO"_3# is limiting;
#4.00cancel("mol LiOH")((1"mol Li"_2"CO"_3)/(2cancel("mol LiOH"))) = color(red)(2.00"mol Li"_2"CO"_3#
#3.20cancel("mol CO"_2)((1"mol Li"_2"CO"_3)/(1cancel("mol CO"_2))) = color(blue)(3.20"mol Li"_2"CO"_3#
We can see that the #4.00"mol LiOH"# produces less #"Li"_2"CO"_3#, so the #"LiOH"# is the limiting reagent, and #color(red)(2.00"mol Li"_2"CO"_3)# will form.
