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Answer 1 · 2016-08-29T01:02:56

Here's what I got.

Explanation:

For part (A), calcium carbonate, ${CaCO}_{3}$ $"CaCO"_3$ , which is insoluble in water, will be dissolved in hydrochloric acid, $HCl$ $"HCl"$ , to form aqueous calcium chloride, ${CaCl}_{2}$ $"CaCl"_2$ , and carbonic acid, $H_{2} {CO}_{3}$ $"H"_2"CO"_3$ .

Because carbonic acid is highly unstable, it will decompose to produce water and release carbon dioxide, ${CO}_{2}$ $"CO"_2$ .

${CaCO}_{3 (s)} + 2 {HCl}_{(a q)} \to {CaCl}_{2 (a q)} + H_{2} O_{(l)} + {CO}_{2 (g)} ↑ ⏐$ $"CaCO"_ (3(s)) + 2"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g)) uarr$

Hydrochloric acid is a strong acid, which means that it dissociates completely in aqueous solution to release hydrogen cations, $H^{+}$ $"H"^(+)$ , and chloride anions, ${Cl}^{-}$ $"Cl"^(-)$ .

${CaCO}_{3 (s)} + 2 \times [H_{(a q)}^{+} + {Cl}_{(a q)}^{-}] \to {Ca}_{(a q)}^{2 +} + 2 {Cl}_{(a q)}^{-} + H_{2} O_{(l)} + {CO}_{2 (g)} ↑ ⏐ ⏐ ⏐$ $"CaCO"_ (3(s)) + 2 xx ["H"_ ((aq))^(+) + "Cl"_ ((aq))^(-)] -> "Ca"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g)) uarr$

The complete ionic equation will look like this

${CaCO}_{3 (s)} + 2 H_{(a q)}^{+} + 2 {Cl}_{(a q)}^{-} \to {Ca}_{(a q)}^{2 +} + 2 {Cl}_{(a q)}^{-} + H_{2} O_{(l)} + {CO}_{2 (g)} ↑ ⏐ ⏐$ $"CaCO"_ (3(s)) + 2"H"_ ((aq))^(+) + 2"Cl"_ ((aq))^(-) -> "Ca"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g)) uarr$

To get the net ionic equation, simply remove the spectator ions from both sides of the equation

$"CaCO"_ (3(s)) + 2"H"_ ((aq))^(+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) -> "Ca"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g)) uarr$

This will get you

$color(green)(|bar(ul(color(white)(a/a)color(black)("CaCO"_ (3(s)) + 2"H"_ ((aq))^(+) -> "Ca"_ ((aq))^(2+) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g)) uarr)color(white)(a/a)|)))$

In part (B), you're dealing with ammonium sulfate, $("NH"_4)_2"SO"_4$ , is a soluble ionic compound that dissociates completely in aqueous solution to form ammonium cations, $"NH"_4^(+)$ , and sulfate anions, $"SO"_4^(2-)$ .

Sodium hydroxide, $"NaOH"$ , is a strong base that dissociates completely to form sodium cations, $"Na"^(+)$ , and hydroxide anions, $"OH"^(-)$ .

When these two solutions are mixed, a neutralization reaction will take place. This reaction will produce aqueous sodium sulfate, $"Na"_2"SO"_4$ , ammonia, $"NH"_3$ , and water.

$("NH"_ 4)_ 2"SO"_ (4(aq)) + 2"NaOH"_ ((aq)) -> "Na"_ 2"SO"_ (4(aq)) + 2"NH"_ (3(aq)) + 2"H"_ 2"O"_ ((l))$

The complete ionic equation is

$2"NH"_ (4(aq))^(+) + "SO"_ (4(aq))^(2-) + 2"Na"_ ((aq))^(+) + 2"OH"_ ((aq))^(-) -> 2"Na"_ ((aq))^(+) + "SO"_ (4(aq))^(2-) + 2"NH"_ (3(aq)) + 2"H"_ 2"O"_ ((l))$

Once again, eliminate the spectator ions

$2"NH"_ (4(aq))^(+) + color(red)(cancel(color(black)("SO"_ (4(aq))^(2-)))) + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + 2"OH"_ ((aq))^(-) -> color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)("SO"_ (4(aq))^(2-)))) + 2"NH"_ (3(aq)) + 2"H"_ 2"O"_ ((l))$

to get the net ionic eqution

$color(green)(|bar(ul(color(white)(a/a)color(black)("NH"_ (4(aq))^(+) + "OH"_ ((aq))^(-) -> "NH"_ (3(aq)) + "H"_ 2"O"_ ((l)))color(white)(a/a)|)))$

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Question #f1eef

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