The chemical reaction between Barium chloride and Na3PO4 takes place as follows:
[tex]3BaCl_{2} + 2Na_{3} PO_{4} ⇒ Ba_{3} (PO_{4})_{2} + 6NaCl[/tex]
2 moles of [tex]Na_{3} PO_{4}[/tex] reacts with 3 moles of [tex]BaCl_{2}[/tex].
Thus, 1.5 moles of [tex]BaCl_{2}[/tex] are required to react with 1 mol of [tex]Na_{3} PO_{4}[/tex] in order for the reaction to be completed.
1.5 moles of BaCl2 are required in order to react completely with 1 mole of Na3PO4.
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