**Basic Information**It takes 6mol of electrons to convert 1mol of KCl to 1mol of KClO

_{3}6mol of electrons = 160.8091AH

1mol of KCl = 74.5513g

1mol of KClO

_{3} = 122.5495g

74.5513g of KCl will yield 122.5495g of KClO

_{3} (at 100% efficiency)

1g of KCl will yield 1.6438g of KClO

_{3}1g of KCl requires 2.1570AH to convert it to KClO

_{3}**Yield Calculations**1)

*Theoretical Yield [g] = Amount of KCl [g] x 1.6438*It is desirable to leave a percentage of KCl in solution to help reduce anode wear and minimise KClO4 production.

2)

*Theoretical Yield [g] = Amount of KCl [g] x (1 - KCl to leave [%]) x 1.6438*The amount of KCl in the cell can be calculated using the cell volume and the solubility of KCl. I chose the solubility value at 20

^{o}C (344g per L) for ease of creating the solution.

3)

*Amount of KCl [g] = Cell Volume [mL] x 0.344*Using 2 & 3

4)

*Theoretical Yield [g] = Cell Volume [mL] x (1 - KCl to leave [%]) x 0.5655*The actual yield will depend on efficiency of the cell which is affected by variables such as pH, temperature, and cell layout.

5)

*Cell Yield [g] = Cell Volume [mL] x (1 - KCl to leave [%]) x Cell Efficiency [%] x 0.5655*

So now that we know how much KClO

_{3} will be produced, how long will it take?

**Run Time Calculations**It takes 160.8091AH to convert 74.5513g of KCl or 2.1570AH per gram. I am going to switch to images for the equations to keep things neat.

6)

Incorporate cell efficiency and percent KCl to leave in solution

7)

Using 3 & 7

8 )

From VK2ZAY, average maximum cell amps is 2A/100mL. We do want to adjust this as different cell designs may tolerate more current, while others may need to be run “cooler”.

9)

Using 8 & 9

10)

Now we can use 5 and 10 as our starting equations

Next we well take in to account some recommendations...