In the reaction between #HCl# and #Na_2CO_3#, what gas is formed?

1 Answer
Dec 14, 2015

Answer:

Carbon dioxide.

Explanation:

Sodium carbonate, #"Na"_2"CO"_3#, will react with hydrochloric acid, #"HCl"#, to produce sodium chloride, a

soluble ionic compound that exists as ions in solution, and carbonic acid, #"H"_2"CO"_3#.

Now, carbonic acid molecules are highly unstable in aqueous solution, so they actually decompose to form carbon dioxide, #"CO"_2#, which bubbles out of solution, and water.

You can thus say that you have

#"Na"_2"CO"_text(3(aq]) + 2"HCl"_text((aq]) -> 2"NaCl"_text((aq]) + "H"_2"CO"_text(3(aq])#

But since

#"H"_2"CO"_text(3(aq]) rightleftharpoons "H"_2"O"_text((l]) + "CO"_text(2(g]) uarr#

Your overall reaction will look like this

#"Na"_2"CO"_text(3(aq]) + 2"HCl"_text((aq]) -> 2"NaCl"_text((aq]) +"H"_2"O"_text((l]) + "CO"_text(2(g]) uarr#

The net ionic equation, for which spectator ions are omitted, will look like this

#"CO"_text(3(aq])^(2-) + 2"H"_text((aq])^(+) -> "H"_2"O"_text((l]) + "CO"_text(2(g]) uarr#