Question #1336b

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Question #1336b

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Answer 1 · 2017-03-05T16:44:44

$H_{2} C_{2} O_{4 (a q)} + Ca {(OH)}_{2 (a q)} \to {CaC}_{2} O_{4 (s)} ⏐ ⏐ ↓ + 2 H_{2} O_{(l)}$ $"H"_ 2"C"_ 2"O"_ (4(aq)) + "Ca"("OH")_ (2(aq)) -> "CaC"_ 2"O"_ (4(s)) darr + 2"H"_ 2"O"_ ((l))$

Explanation:

The unbalanced chemical equation should look like this

$H_{2} C_{2} O_{4 (a q)} + Ca {(OH)}_{2 (a q)} \to {CaC}_{2} O_{4 (s)} ⏐ ⏐ ↓ + H_{2} O_{(l)}$ $"H"_ 2"C"_ 2"O"_ (4(aq)) + "Ca"("OH")_ (2(aq)) -> "CaC"_ 2"O"_ (4(s)) darr + "H"_ 2"O"_ ((l))$

This reaction has oxalic acid, $H_{2} C_{2} O_{4}$ $"H"_2"C"_2"O"_4$ , and calcium hydroxide, $Ca {(OH)}_{2}$ $"Ca"("OH")_2$ , as the reactants and calcium oxalate, ${CaC}_{2} O_{4}$ $"CaC"_2"O"_4$ , and water as the products.

Calcium hydroxide is not very soluble in water, but the amount that does dissolve dissociates completely to produce calcium cations, ${Ca}^{2 +}$ $"Ca"^(2+)$ , and hydroxide anions, ${OH}^{-}$ $"OH"^(-)$ .

You can thus say that you're dealing with a neutralization reaction between a weak acid and a strong base.

You can balance this equation by taking a look at the ions involved in the reaction.

$2 H_{(a q)}^{+} + {C_{2} O_{4}}_{(a q)}^{2 -} + {Ca}_{(a q)}^{2 +} + 2 {OH}_{(a q)}^{-} \to Ca {C_{2} O_{4}}_{(s)} ⏐ ⏐ ↓ + H_{2} O_{(l)}$ $2"H"_ ((aq))^(+) + color(blue)("C"_ 2"O"_ 4)_ ((aq))^(color(blue)(2-)) + color(red)("Ca")_ ((aq))^(color(red)(2+)) + 2"OH"_ ((aq))^(-) -> color(red)("Ca")color(blue)("C"_ 2"O"_ 4) _ ((s)) darr + "H"_ 2"O"_ ((l))$

Notice that the $1$ $1$ calcium cation and $1$ $1$ oxalate anion, $C_{2} O_{4}^{2 -}$ $"C"_2"O"_4^(2-)$ , present on the reactants' side are accounted for on the products' side.

This means that all you have to do in order to balance this chemical equation is to balance the hydrogen and oxygen atoms.

Excluding the aforementioned oxalate anion, you have

$4 \times H$ $4 xx "H"$ on the reactants' side $\to 2 H^{+} + 2 {OH}^{-}$ $-> 2"H"^(+) + 2"OH"^(-)$

$2 \times H$ $2 xx "H"$ on the products' side $\to H_{2} O$ $-> "H"_2"O"$

and

$2 \times O$ $2 xx "O"$ on the reactants 'side $\to 2 {OH}^{-}$ $-> 2"OH"^(-)$

$1 \times O$ $1 xx "O"$ on the products' side $\to H_{2} O$ $-> "H"_2"O"$

You can thus balance the hydrogen and oxygen atoms by adding a coefficient of $2$ $2$ to the water molecule.

The balanced chemical equation will thus be

$color(darkgreen)(ul(color(black)("H"_ 2"C"_ 2"O"_ (4(aq)) + "Ca"("OH")_ (2(aq)) -> "CaC"_ 2"O"_ (4(s)) darr + 2"H"_ 2"O"_ ((l)))))$

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Question #1336b

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