Balance the following equation with the smallest whole number coefficients. Coefficients of one must be entered to received credit. Li2CO3 (aq) + SnCl4 (aq) ⟶ LiCl (aq) +   Sn(CO3)2 (s)

To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation.

1. First, let's balance the Lithium (Li) atoms. We have 2 Li atoms on the left side and 1 on the right side. To balance this, we put a coefficient of 2 in front of LiCl on the right side. Now, the equation looks like this:

   Li2CO3 (aq) + SnCl4 (aq) ⟶ 2LiCl (aq) + Sn(CO3)2 (s)

2. Next, let's balance the Chlorine (Cl) atoms. We have 4 Cl atoms on the left side and 2 on the right side. To balance this, we put a coefficient of 2 in front of SnCl4 on the left side. Now, the equation looks like this:

   Li2CO3 (aq) + 2SnCl4 (aq) ⟶ 2LiCl (aq) + Sn(CO3)2 (s)

3. Now, let's balance the Tin (Sn) atoms. We have 2 Sn atoms on the left side and 1 on the right side. To balance this, we put a coefficient of 2 in front of Sn(CO3)2 on the right side. Now, the equation looks like this:

   Li2CO3 (aq) + 2SnCl4 (aq) ⟶ 2LiCl (aq) + 2Sn(CO3)2 (s)

4. Finally, let's balance the Carbon (C) and Oxygen (O) atoms. We have 1 C atom and 3 O atoms on the left side and 2 C atoms and 6 O atoms on the right side. To balance this, we put a coefficient of 2 in front of Li2CO3 on the left side. Now, the equation looks like this:

   2Li2CO3 (aq) + 2SnCl4 (aq) ⟶ 2LiCl (aq) + 2Sn(CO3)2 (s)

Now, the equation is balanced with the smallest whole number coefficients.