If 0.5 mol of BaCl2 is mixed with 0.20 mol of Na3PO4, the maximum amount of Ba3(PO4)2 that can be formed is(a) 0.70 mol (b) 0.50 mol (c) 0.20 mol (d) 0.1 molplease explain how?

I think it''s a mole -1 chptr. qstn.

             3BaCl2 + 2Na3PO4  ------------->  Ba3(PO4)2 + 6Nacl

                  0.5 mol           0.20 mol               x mol           

                First find which is the limiting reagent?

              0.5/3 = 1/6          0.2/2 = 1/10

                            1/10 < 1/6

              thus,  Na3PO4 is the limiting reagent.

               Now, 1/10 = x/1

                             x = 0.1 mol.

(d) may be the ansr.

plz approvE!

Approved

3 BaCl2 + 2 Na3PO4 = Ba3(PO4)2 + 6NaCl
Here clearly BaCl2 is the limiting agent
So 3 moles of bacl2 gives 1 mole of ba3(po4)2
so 0.5 mole will give 0.2 mole(approx)

I hope i am correct 

3bacl2 +2na3po4=ba3(po4)2+6naclHear clearly NA3PO4 is limiting reagentNow 1/10=x/1 X=0.1molSo D is the answer

Dear Student,
Please find below the solution to your problem.

3 BaCl2 + 2 Na3PO4 -------------> Ba3(PO4)2 + 6 NaCl
0.5 mol 0.20 mol x mol
Lets first find the limiting reagent?
0.5/3 = 1/6 0.2/2 = 1/10
1/10 < 1/6
thus, Na3PO4 is the limiting reagent.
Now, 1/10 = x/1
x = 0.1 mol.

Thanks and Regards