How do you balance #Na_2SO_4 + Ca(NO_3)_2 -> CaSO_4 + NaNO_3#?
1 Answer
Explanation:
Every time you're dealing with a double replacement reaction, you can make your life easier by looking at ions, rather than at atoms, when trying to balance the chemical equation.
A double replacement reaction is characterized by the reaction of the soluble ionic compounds and the formation of an insoluble solid that precipitates out of solution.
Sodium sulfate,
#"Na"_2"SO"_text(4(aq]) -> color(red)(2)"Na"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#
#"Ca"("NO"_3)_text(2(aq]) -> "Ca"_text((aq])^(2+) + color(blue)(2)"NO"_text(3(aq])^(-)#
The reaction produces calcium sulfate, an insoluble solid that precipitates out of solution, and sodium nitrate,
#"NaNO"_text(3(aq]) -> "Na"_text((aq])^(+) + "NO"_text(3(aq])^(-)#
Now, the idea here is that all the ions present on the reactants' side must be present, either as dissociated ions or as part of the solid, on the products' side.
Notice that the one calcium cation,
On the other hand, notice that you have
This tells you that will have to multiply the sodium nitrate by
The complete chemical equation for this reaction will look like this
#"Na"_2"SO"_text(4(aq]) + "Ca"("NO"_3)_text(2(aq]) -> "CaSO"_text(4(s]) darr + color(green)(2)"NaNO"_text(3(aq])#
The net ionic equation, which doesn't include spectator ions, i.e. the ions that exist dissociated on both sides of the equation, will look like this
#"SO"_text(4(aq])^(2-) + "Ca"_text((aq])^(2+) -> "CaSO"_text(4(s]) darr#