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Question

If 0.5 moles of BaCl2 is mixed with 0.2 moles of Na3PO4 the maximum moles of Ba3(PO4)2 obtained is

A
0.2
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B
0.5
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C
0.3
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D
0.1
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Solution

The correct option is D 0.1
3BaCl2+2Na3PO4Ba3(PO4)2+6NaCl


Finding the limiting reagent:
given moles of BaCl2 stoichiometric coefficient=0.53

given moles of Na3PO4 stoichiometric coefficient=0.22

Hence the limiting reagent is Na3PO4

From stoichiometry,
2 moles of Na3PO4 gives 1 mole of Ba3(PO4)2
So, 0.2 mole give 0.1 mole of Ba3(PO4)2

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