What is the net ionic equation for lithium carbonate and aluminum nitrate?

Question

lithium carbonate: Li₂CO₃

aluminum nitrate: Al(NO₃)₃

Answer 1

Li₂CO₃ + Al(NO₃)₃

The first thing to do is find the products. To do this, separate the lithium carbonate and aluminum nitrate into their ions:

Li⁺ CO₃^-2 + Al⁺³ NO₃^-

Then switch the metals and cross your charges:

Al⁺³ CO₃^-2 + Li⁺ NO₃^- ==== Al₂(CO₃)₃ + LiNO₃

So your equation with products is:

Li₂CO_{3 (aq)} + Al(NO₃)_{3 (aq)} -- > Al₂(CO₃)₃ + LiNO₃

Now balance the equation and determine which of the products will precipitate.

To look at solubility rules, any group 1 element with nitrate will always be aqueous, so the aluminum carbonate will be the solid.

Li₂CO_{3 (aq)} + Al(NO₃)_{3 (aq)} -- > Al₂(CO₃)_{3 (s)} + LiNO_{3 (aq)}

To balance the equation, I always suggest making a chart with each element or polyatomic ion and the amount on the reactants and products sides:

Reactants Products

Li 2 1

CO₃ 1 3

NO₃ 3 1

Al 1 2

Given the chart, we will want to put a 6 in front of the LiNO₃ , 3 in front of the Li₂CO₃, and 2 in front of the Al(NO₃)₃ which will give us:

Reactants Products

Li 6 6

CO₃ 3 3

NO₃ 6 6

Al 2 2

Therefore our balanced equation is:

3 Li₂CO_{3 (aq)} + 2 Al(NO₃)_{3 (aq)} -- > Al₂(CO₃)_{3 (s)} + 6 LiNO_{3 (aq)}

Now we will ionize everything that is aqueous. Keep the solid together!

6 Li⁺(aq) + 3 CO₃^-2 (aq) + 2 Al⁺³ (aq) + 6 NO₃^- (aq) --> Al₂(CO₃)₃ (s) + 6 Li⁺ (aq) + 6 NO₃^- (aq)

Cancel the ions shown on the reactants and products sides:

6 Li⁺(aq) + 3 CO₃^-2 (aq) + 2 Al⁺³ (aq) + 6 NO₃^- (aq) --> Al₂(CO₃)₃ (s) + 6 Li⁺ (aq) + 6 NO₃^- (aq)

Now just pull down what's left and that's your net ionic equation:

3 CO₃^-2 (aq) + 2 Al⁺³ (aq) --> Al₂(CO₃)₃ (s)