Question #f1eef

1 Answer
Aug 29, 2016

Answer:

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)|)))#