Question #44dbb

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Question #44dbb

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Answer 1 · 2015-06-28T23:59:24

You can balance this double replacement reaction by inspection.

Explanation:

The unbalanced chemical equation looks like this

$N a_{3} P O_{4 (a q)} + B a {(N O_{3})}_{2 (a q)} \to N a N O_{3(aq]} + B a_{3} {(P O_{4})}_{2 (s)}$ $Na_3PO_(4(aq)) + Ba(NO_3)_(2(aq)) -> NaNO_text(3(aq]) + Ba_3(PO_4)_(2(s))$

The easiest way to balance this equation is to treat the phosphate, $P O_{4}^{3 -}$ $PO_4^(3-)$ , and nitrate, $N O_{3}^{-}$ $NO_3""^(-)$ , polyatomic ions as a single unit.

So, notice that you have 2 phosphate ions on the products' side, but only one on the reactants' side. Multiply sodium phosphate by 2 to balance the phosphate ions.

$2 N a_{3} P O_{4 (a q)} + B a {(N O_{3})}_{2 (a q)} \to N a N O_{3(aq]} + B a_{3} {(P O_{4})}_{2 (s)}$ $color(red)(2)Na_3PO_(4(aq)) + Ba(NO_3)_(2(aq)) -> NaNO_text(3(aq]) + Ba_3(PO_4)_(color(red)(2)(s))$

Now focus on the nitrate ions. Notice that you have 2 nitrate ions on the reactants' side, but only 1 on the products' side. Multiply sodium nitrate by 2 to balance the nitrate ions.

$2 N a_{3} P O_{4 (a q)} + B a {(N O_{3})}_{2 (a q)} \to 2 N a N O_{3(aq]} + B a_{3} {(P O_{4})}_{2 (s)}$ $color(red)(2)Na_3PO_(4(aq)) + Ba(NO_3)_(2(aq)) -> color(green)(2)NaNO_text(3(aq]) + Ba_3(PO_4)_(color(red)(2)(s))$

Now focus on the sodium and barium atoms. Notice that you have 6 sodium atoms on the reactants' side, but only 2 on the products' side $\to$ $->$ multiply the sodium nitrate by 3 to balance the sodium atoms.

$2 N a_{3} P O_{4 (a q)} + B a {(N O_{3})}_{2 (a q)} \to 6 N a N O_{3(aq]} + B a_{3} {(P O_{4})}_{2 (s)}$ $color(red)(2)Na_3PO_(4(aq)) + Ba(NO_3)_(2(aq)) -> color(green)(6)NaNO_text(3(aq]) + Ba_3(PO_4)_(color(red)(2)(s))$

Now the nitrate ions are unbalanced. Multiply the barium nitrate by 3 to balance the 6 nitrate ions on the products' side with the those on the reactants' side.

$2 N a_{3} P O_{4 (a q)} + 3 B a {(N O_{3})}_{2 (a q)} \to 6 N a N O_{3(aq]} + B a_{3} {(P O_{4})}_{2 (s)}$ $color(red)(2)Na_3PO_(4(aq)) + color(blue)(3)Ba(NO_3)_(2(aq)) -> color(green)(6)NaNO_text(3(aq]) + Ba_3(PO_4)_(color(red)(2)(s))$

As it turns out, the barium atoms are balanced, 3 on the reactants' side and 3 on the products' side.

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Question #44dbb

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