Question

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?

Answer 1

`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.

Answer 2

`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