The maximum amount of ba3(po4)2, Chemistry

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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

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

 


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