Write the Gross and Net Ionic Equations for the following Reactions:
LiNO3(aq)+(NH4)3N(aq) => Li3N(s)+ NH4NO3(aq)
Al(s)+CuCl2(aq) => AlCl3(aq)+Cu(s)
5 answers
And you are having trouble doing what?
Yep
C'mon. Yep won't get it. Exactly what do you not understand?
I'm not sure what it's asking. I know I need to write out the aqueous parts and write the charges, but I don't how to do that. Mostly I need help getting it to the Gross Equation.
First you balance the molecular equation.
3LiNO3(aq)+(NH4)3N(aq) => Li3N(s)+ 3NH4NO3(aq)
Then convert to a total ionic equation following a few rules as follows:
If the compound is an ionic compound you break it apart into ions. If it is a strong acid or strong base do likewise. If a solid, a gas, or if a weak acid or weak base, write those as molecules.
3Li^+(aq) + 3NO3^-(aq) + 3NH4^+ + N^3-(aq) => Li3N(s)+ 3NH4^+(aq) + 3NO3^-(aq)
That is the total ionic equation. Now you look for ions that are common to both sides; cancel those you find. I see 3 NO3^- on the left and right and I see 3 NH4^+ left and right sides. Cancel those from both sides and you are left with this.
3Li^+(aq) + N^3-(aq) ==> Li3N(s)
That is the net ionic equation. The ions you canceled are called spetator ions; they don't enter into the reaction. I'll be happy to check your work on the next one.
3LiNO3(aq)+(NH4)3N(aq) => Li3N(s)+ 3NH4NO3(aq)
Then convert to a total ionic equation following a few rules as follows:
If the compound is an ionic compound you break it apart into ions. If it is a strong acid or strong base do likewise. If a solid, a gas, or if a weak acid or weak base, write those as molecules.
3Li^+(aq) + 3NO3^-(aq) + 3NH4^+ + N^3-(aq) => Li3N(s)+ 3NH4^+(aq) + 3NO3^-(aq)
That is the total ionic equation. Now you look for ions that are common to both sides; cancel those you find. I see 3 NO3^- on the left and right and I see 3 NH4^+ left and right sides. Cancel those from both sides and you are left with this.
3Li^+(aq) + N^3-(aq) ==> Li3N(s)
That is the net ionic equation. The ions you canceled are called spetator ions; they don't enter into the reaction. I'll be happy to check your work on the next one.
4 answers