Balance the following equation:

Here are instructions for balancing redox equations. Here are some hints to get you started. Cr changes from +6 for each atom on the left to +3 on the right.

S changes from +4 on the left to +6 on the right.
http://www.chemteam.info/Redox/Redox.html

1,2,3,4

To balance the given equation:

K2CrO4 + Na2SO3 + HCl >> Na2SO4 + CrCl3 + H2O

The coefficients for each compound can be determined as follows:

2 K2CrO4 + 3 Na2SO3 + 8 HCl >> 3 Na2SO4 + 2 CrCl3 + 7 H2O

Therefore, the balanced equation is:

2 K2CrO4 + 3 Na2SO3 + 8 HCl >> 3 Na2SO4 + 2 CrCl3 + 7 H2O

To balance the equation K2CrO4 + Na2SO3 + HCl >>> Na2SO4 + CrCl3 + H2O, we need to make sure that there is an equal number of each type of atom on both sides of the equation.

Let's go through the process of balancing the equation step by step:

Step 1: Start with the atoms that appear in only one compound on each side of the equation.

- There are 2 K atoms on the left side and 0 K atoms on the right side. So, we will put a coefficient of 2 in front of KCl to balance K atoms.

2K2CrO4 + Na2SO3 + 2HCl >>> Na2SO4 + CrCl3 + H2O

- There are 4 O atoms on the left side and 4 O atoms on the right side. So, O atoms are already balanced.

2K2CrO4 + Na2SO3 + 2HCl >>> Na2SO4 + CrCl3 + H2O

Step 2: Now let's move on to the next atom(s) that are not balanced.

- There are 2 Cr atoms on the left side and 1 Cr atom on the right side. To balance these, we will put a coefficient of 2 in front of CrCl3.

2K2CrO4 + Na2SO3 + 2HCl >>> Na2SO4 + 2CrCl3 + H2O

- There is 1 Na atom on the left side and 2 Na atoms on the right side. To balance these, we will put a coefficient of 2 in front of Na2SO4.

2K2CrO4 + Na2SO3 + 2HCl >>> 2Na2SO4 + 2CrCl3 + H2O

- There are 3 Cl atoms on the left side and 6 Cl atoms on the right side. To balance these, we will put a coefficient of 3 in front of HCl.

2K2CrO4 + Na2SO3 + 3HCl >>> 2Na2SO4 + 2CrCl3 + H2O

Step 3: Finally, let's check if the remaining atoms are balanced.

- There are 2 S atoms on the left side and 2 S atoms on the right side. So, S atoms are balanced.

2K2CrO4 + Na2SO3 + 3HCl >>> 2Na2SO4 + 2CrCl3 + H2O

- There are 6 H atoms on the left side and 6 H atoms on the right side. So, H atoms are balanced.

2K2CrO4 + Na2SO3 + 3HCl >>> 2Na2SO4 + 2CrCl3 + 3H2O

Now we have balanced the equation:

2K2CrO4 + Na2SO3 + 3HCl >>> 2Na2SO4 + 2CrCl3 + 3H2O

The coefficients for each compound, separated by commas in the order they appear, are:

2, 1, 3, 2, 2, 3.