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Sagot :
Let's rewrite the unbalanced equation:
[tex]Ba(C_2H_3O_2)_2+Na_3PO_4\to Ba_3(PO_4)_2+NaC_2H_3O_2[/tex]For the equation to be balanced, we need all the atoms to have equal number in each side.
However, we have som unchanged groups: C₂H₃O₂⁻ and PO₄³⁻ are unmodified, so we can consider them as one thing.
So, starting from Ba, we see that we have 1 on the left side and 3 on the right side, so let's start by adding a coefficient 3 to the left side so that Ba gets balanced:
[tex]3Ba(C_2H_3O_2)_2+Na_3PO_4\to Ba_3(PO_4)_2+NaC_2H_3O_2[/tex]Now, the group C₂H₃O₂⁻ appear 6 times in the left side and only once on the right side. We can fix that by adding a coefficient 6 to NaC₂H₃O₂:
[tex]3Ba(C_2H_3O_2)_2+Na_3PO_4\to Ba_3(PO_4)_2+6NaC_2H_3O_2[/tex]Now, we can look to Na: we have 3 on the left side and 6 on the right side. So, we can add a coefficient 2 on Na₃PO₄:
[tex]3Ba(C_2H_3O_2)_2+2Na_3PO_4\to Ba_3(PO_4)_2+6NaC_2H_3O_2[/tex]Now we check wether PO₄³⁻ is balanced: it appears twice on the left side and twice on the right side, so it is balanced.
Thus, the balanced equation is:
[tex]3Ba(C_2H_3O_2)_2+2Na_3PO_4\to Ba_3(PO_4)_2+6NaC_2H_3O_2[/tex]Also, we can see that the group C₂H₃O₂⁻ went from Ba to Na, and that the group PO₄³⁻ went from Ba to Na, that is both groups were displaced.
Thus, this is a double-displacement reaction.
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