Write a balanced equation, ionic equation and the net ionic equation for the following:

A) Li2SO4 + Sr(NO3)3 --->
B) H2SO4 + Na2CO3 --->

You made a typo: that's Sr(NO3)2

Molecular:
Li2SO4(aq) + Sr(NO3)2(aq) ==> SrSO4(s) + 2LiNO3(aq)

ionic:
2Li^+(aq) + SO4^2-(aq) + Sr^2+(aq) + 2 NO3^-(aq) ==> SrSO4(s) + 2Li^+(aq) + 2NO3^-(aq)

NET ionic:
Just cancel the ions common to both sides to leave you with
Sr^2+(aq) + SO4^2-(aq) ==> SrSO4(s)

A) Li2SO4 + Sr(NO3)3 ---> 2LiNO3 + SrSO4

B) H2SO4 + Na2CO3 ---> H2CO3 + Na2SO4

Now, let's break it down:

A) Ionic equation:
2Li⁺ + SO4²⁻ + Sr²⁺ + 3NO3⁻ ---> 2Li⁺ + 2NO3⁻ + SrSO4

B) Ionic equation:
2H⁺ + SO4²⁻ + 2Na⁺ + CO3²⁻ ---> H2CO3 + 2Na⁺ + SO4²⁻

And finally, the net ionic equations:

A) 2Li⁺ + Sr²⁺ ---> 2Li⁺ + SrSO4

B) 2H⁺ + CO3²⁻ ---> H2CO3

A) Li2SO4 + Sr(NO3)3 -> 2LiNO3 + SrSO4

Balanced Equation:
2Li2SO4 + 3Sr(NO3)3 -> 6LiNO3 + SrSO4

Ionic Equation:
2Li⁺ + SO4²⁻ + 3Sr²⁺ + 6NO3⁻ -> 6Li⁺ + 6NO3⁻ + SrSO4

Net Ionic Equation:
2Li⁺ + SO4²⁻ + 3Sr²⁺ -> SrSO4 + 2Li⁺

B) H2SO4 + Na2CO3 -> H2O + Na2SO4 + CO2
Balanced Equation:
H2SO4 + Na2CO3 -> H2O + Na2SO4 + CO2

Ionic Equation:
2H⁺ + SO4²⁻ + 2Na⁺ + CO3²⁻ -> 2H2O + 2Na⁺ + SO4²⁻ + CO2

Net Ionic Equation:
2H⁺ + CO3²⁻ -> H2O + CO2

A) Li2SO4 + Sr(NO3)3 --->

To write the balanced equation, you need to first determine the formulas of the products formed. Here's how you can do it:

1. Write the formulas of the reactants:
Lithium sulfate (Li2SO4)
Strontium nitrate (Sr(NO3)3)

2. Balance the equation by making sure the number of atoms of each element is the same on both sides:
3Li2SO4 + Sr(NO3)3 --->

3. Identify the products by combining the cation of one reactant with the anion of the other reactant:
Lithium sulfate contains the cation Li+ and the anion SO4^2-
Strontium nitrate contains the cation Sr^2+ and the anion NO3-

4. Write the ionic equation by representing the reactants and products as dissociated ions:
3Li^+ + 3SO4^2- + Sr^2+ + 6NO3- --->

5. Write the net ionic equation by removing the spectator ions that appear on both sides of the equation:
The spectator ions are NO3-, so they are removed from the net ionic equation:
3Li^+ + 3SO4^2- + Sr^2+ ---> 2Li^+ + 2SO4^2- + Sr^2+

Therefore, the balanced equation is:
3Li2SO4 + Sr(NO3)3 ---> 2LiNO3 + SrSO4

B) H2SO4 + Na2CO3 --->

Follow the same steps as in the previous equation:

1. Write the formulas of the reactants:
Sulfuric acid (H2SO4)
Sodium carbonate (Na2CO3)

2. Balance the equation:
H2SO4 + Na2CO3 --->

3. Identify the products:
Sulfuric acid contains the cation H+ and the anion SO4^2-
Sodium carbonate contains the cation Na+ and the anion CO3^2-

4. Write the ionic equation:
2H+ + SO4^2- + 2Na+ + CO3^2- --->

5. Write the net ionic equation by removing the spectator ions (2Na+ and SO4^2-):
2H+ + CO3^2- ---> H2CO3

Therefore, the balanced equation is:
H2SO4 + Na2CO3 ---> Na2SO4 + H2CO3