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# 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 mol please explain how?

Vikas TU
14149 Points
8 years ago

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!

Roshan Mohanty
64 Points
8 years ago

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

Saksham Soni
26 Points
4 years ago
3bacl2 +2na3po4=ba3(po4)2+6naclHear clearly NA3PO4 is limiting reagentNow 1/10=x/1 X=0.1molSo D is the answer
Rishi Sharma
one year ago
Dear Student,

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