Question #38d7f

1 Answer
Apr 14, 2016


#"Mg"_ ((aq))^(2+) + "CO"_ (3(aq))^(2-) -> "MgCO"_ (3(s))#


You're mixing magnesium nitrate, #"Mg"("NO"_3)_2#, and sodium carbonate, #"Na"_2"CO"_3#, two soluble ionic compounds that dissociate completely in aqueous solution to form cations and anions.

You can thus represent the two reactants as

#"Mg"("NO"_ 3)_ (2(aq)) -> "Mg"_ ((aq))^(2+) + 2"NO"_(3(aq))^(-)#

#"Na"_ 2"CO"_ (3(aq)) -> 2"Na"_ ((aq))^(+) + "CO"_(3(aq))^(2-)#

Notice that the reaction produces magnesium carbonate, #"MgCO"_3#, an insoluble solid that precipitates out of solution, and aqueous sodium nitrate, #"NaNO"_3#, which will exist as sodium cations and nitrate anions in solution.

This means that you can rewrite the balanced chemical equation that describes this double replacement reaction like this

#"Mg"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2"Na"_ ((aq))^(+) + "CO"_ (3(aq))^(2-) -> "MgCO"_ (3(s)) darr + 2 xx ["Na"_ ((aq))^(+) + "NO"_(3(aq))^(-)]#

This represents the complete ionic equation. To get the net ionic equation, you need to remove spectator ions, i.e. ions that are present on both sides of the equation.

In your case, you'll have

#"Mg"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + "CO"_ (3(aq))^(2-) -> "MgCO"_ (3(s)) darr + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-))))#

which is equivalent to

#color(green)(|bar(ul(color(white)(a/a)color(black)("Mg"_ ((aq))^(2+) + "CO"_ (3(aq))^(2-) -> "MgCO"_ (3(s)) darr)color(white)(a/a)|)))#

It's worth noting that magnesium carbonate is a white precipitate.