Question #65be7

1 Answer

Answer:

2#Ag^+#(aq) + #CO_3^-2#(aq) -> #Ag_2CO_3#(s)

Explanation:

The steps involved are:
* Write the skeleton equation

#AgNO_3#(aq) + #Na_2CO_3#(aq) -> #Ag_2CO_3#(s) + #NaNO_3#(aq)
note: you know the silver carbonate is the precipitate(s) because nitrate compounds are always water soluble

*Write the (aq) compounds show show the ions have been pulled apart by the dissolving properties of water

#Ag^+#(aq) + #NO_3^-#(aq) + #Na^+#(aq) + #CO_3^-2#(aq) ->
#Ag_2CO_3#(s) + #Na^+#(aq) + #NO_3^-#(aq)

*Identify spectator ions which exist in the dissolved state as both reactants and products and cross them out then re-write the remaining ions and precipitated compound.
(#Na^+#(aq) and #NO_3^-#(aq) are the spectator ions)

#Ag^+#(aq) + #CO_3^-2#(aq) -> #Ag_2CO_3#(s)

Now balance number of ions and charges to find the final answer shown at the top.

Here is a video which gives another example of how to write a net ionic equation for a precipitation reaction.

Hope this helps!