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 BaCl₂is the limiting agent

So 3 moles of bacl₂gives 1 mole of ba₃(po4)₂

so 0.5 mole will give 0.2 mole

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