AP Chem ---

PbS + O2 = PbO + SO2
Balance the equation and write a short paragraph explaining the electron transfers that happen.

Pb is +2 on the left and +2 on the right

O is zero on the left and -2 on the right.
S is -2 on the left and +4 on the right.

To balance the equation PbS + O2 = PbO + SO2, we need to ensure that the same number of atoms of each element are present on both sides of the equation. Here's how you can balance the equation:

1. Start by counting the number of atoms for each element on both sides of the equation.

On the left-hand side: Pb (1), S (1), O (2)
On the right-hand side: Pb (1), O (1), S (1), O (2)

2. Balance the atoms one element at a time, starting with the element that appears in the fewest molecules. In this case, it is the Sulfur (S).

Balance the sulfur atoms by placing a coefficient of 1 in front of PbS: PbS + O2 = PbO + SO2.

3. Next, balance the oxygen atoms. There are 2 oxygen atoms on the left side and 3 on the right side.

To balance the oxygen atoms, place the coefficient 2 in front of O2: PbS + 2O2 = PbO + SO2.

The balanced equation is now: PbS + 2O2 = PbO + SO2.

As for the electron transfers that occur in this reaction, it involves oxidation and reduction. In this equation, sulfur (S) undergoes oxidation, while lead (Pb) undergoes reduction. During the reaction, sulfur gains oxygen atoms, resulting in the formation of sulfur dioxide (SO2), which is the oxidized form of sulfur. On the other hand, lead loses oxygen atoms, forming lead oxide (PbO), which is the reduced form of lead. These oxidation and reduction reactions involve the transfer of electrons between the reacting species.