How do you balance ___AgNO3+___(NH4)2CrO4>___Ag2CrO4+___NH4NO3?

1 Answer
Oct 18, 2015

Answer:

#2"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_2"CrO"_text(4(s]) darr + 2"NH"_4"NO"_text(3(aq])#

Explanation:

Start by writing down the unbalanced chemical equation for this double replacement reaction

#"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_2"CrO"_text(4(s]) darr + "NH"_4"NO"_text(3(aq])#

In this reaction, silver nitrate, #"AgNO"_3#, will react with ammonium chromate, #("NH"_4)_2"CrO"_4#, both soluble compounds, to form Silver chromate, #"Ag"_2"CrO"_4#, an insoluble solid, and ammonium nitrate, #"NH"_4"CO"_3#, also a soluble compound.

When you're dealing with ionic compounds, you can treat cations and anions as units. This will allow you to balance the chemical equation faster than if you went by atoms alone.

So, the cations are

  • #"Ag"^(+)#
  • #"NH"_4^(+)#

and the anions are

  • #"NO"_3^(-)#
  • #"CrO"_4^(2-)#

So, start with the silver cations. Notice that you have one silver cation on the reactants' side and two on the products' side. Multiply the silver nitrate by #2# to get

#color(red)(2)"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_color(red)(2)"CrO"_text(4(s]) darr + "NH"_4"NO"_text(3(aq])#

Now look at the ammonium cations. You have two ammonium cations on the reactants' side, and only one on the products' side.

Multiply the ammonium nitrate by #2# to get

#color(red)(2)"AgNO"_text(3(aq]) + ("NH"_4)_color(green)(2)"CrO"_text(4(aq]) -> "Ag"_color(red)(2)"CrO"_text(4(s]) darr + color(green)(2)"NH"_4"NO"_text(3(aq])#

The nitrate aniuons are balanced, since you have two on the products' side and two on the reactants' side.

The same can be said for the chromate anions, since you have one on the reactants's side and one on the products' side.

Therefore, the balanced chemical equation for this reaction is

#2"AgNO"_text(3(aq]) + ("NH"_4)_2"CrO"_text(4(aq]) -> "Ag"_2"CrO"_text(4(s]) darr + 2"NH"_4"NO"_text(3(aq])#