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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?
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!
3 BaCl2 + 2 Na3PO4 = Ba3(PO4)2 + 6NaClHere clearly BaCl2 is the limiting agentSo 3 moles of bacl2 gives 1 mole of ba3(po4)2so 0.5 mole will give 0.2 mole(approx)I hope i am correct
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