Silver Dichromate Balanced Equation

The balanced equation for the formation of silver dichromate is:

2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3

In this reaction, silver nitrate (AgNO3) reacts with sodium chromate (Na2CrO4) to form silver dichromate (Ag2CrO4) and sodium nitrate (NaNO3).

To balance the equation, we need to make sure that the same number of atoms of each element are present on both sides of the equation.

The coefficients in front of each compound indicate the number of molecules or formula units of each compound involved in the reaction. In this case, we need to balance the number of silver, chromium, oxygen, nitrogen, and sodium atoms.

On the left-hand side of the equation, there are two silver atoms, two nitrogen atoms, one chromium atom, four oxygen atoms, and two sodium atoms. On the right-hand side of the equation, there are two silver atoms, one chromium atom, four oxygen atoms, and two sodium atoms.

To balance the equation, we can start by ensuring that the same number of chromium atoms are present on both sides. To do this, we add a coefficient of 2 in front of Na2CrO4:

2AgNO3 + 2Na2CrO4 → Ag2CrO4 + 2NaNO3

Now we have two chromium atoms on both sides, but the number of sodium atoms has increased to four on the left-hand side. To balance this, we add a coefficient of 2 in front of NaNO3:

2AgNO3 + 2Na2CrO4 → Ag2CrO4 + 4NaNO3

Now the number of sodium atoms is equal on both sides, but the number of nitrogen atoms has increased to four on the right-hand side. We can balance this by adding a coefficient of 2 in front of AgNO3:

4AgNO3 + 2Na2CrO4 → 2Ag2CrO4 + 4NaNO3

Finally, we have a balanced equation with four silver atoms, two chromium atoms, eight oxygen atoms, four nitrogen atoms, and four sodium atoms on both sides.