What is the limiting reactant when 6.00 g of BaCl2 reacts with 5.00 g of Na3PO4 to form the precipitate Ba3(PO4)2?

To find out the limiting reactant, you calculate the number of moles of the reactant and the limiting reactant is the one with the least number of moles.
The reaction is;
#3BaCl_2 + 2Na_3PO_4 -> Ba_3 (PO_4) _2 + 6NaCl #

No. of moles = #(mass)/(Mr)#
In this case,

No. of moles ( #Na_3PO_4#) = #(5.00)/601.93#= 0.00831 moles
No. of moles of (#BaCl_2#) = #(6.00/208.23)#=0.0288 moles
This number of moles if for 1 molecule of the reactants.

But in this case 3 molecules of #BaCl_2# reacts with 2 molecules of #Na_3PO_4 #,

No. of moles of 3 molecules of #BaCl_2#= #0.0288*3#=0.0864moles.
No. of moles of 2 molecules of #Na_3PO_4 #=#0.00831*2#=0.0166 moles.

So, as #Na_3PO_4 # has less number of moles than #BaCl_2#, #Na_3PO_4 # acts as the limiting reactant.

you have to balance the equation
#3BaCl_2 + 2Na_3PO_4 = Ba_3(PO_4)_2 + 6 NaCl#
from the ratio of reaction you see that are need 3 mol of #BaCl_2# for 2 mol of#Na_3PO_4# to give one mol of# Ba_3(PO_4)_2#
Then you have to calculate the mol of the reactants
#mol = (Mg)/ (MM)#
#mol BaCl_2 = 6,00/208,23=0,0288 mol#
#mol Na_3PO_4 = 5,00/ 163,94 = 0,0305 mol#
You have less #BaCl_2#. This is the limitant reactant. It can react only 2/3 0,0288 of the mol of #Na_3PO_4# =0,0192 mol and you obtain 1/3 0,0288 mol of# Ba_3(PO_4)_2# =0,0960