Question

Question #f1eef

Answer 1

Here's what I got.

Explanation:

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

Because carbonic acid is highly unstable, it will decompose to produce water and release carbon dioxide, `"CO"_2`.

`"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"^(+)`, and chloride anions, `"Cl"^(-)`.

`"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"_ ((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)|)))`