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?

2 Answers
May 7, 2017

Answer:

#Na_3PO_4 #

Explanation:

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.

May 7, 2017

Answer:

#BaCl_2#

Explanation:

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