A holding tank requires chlorination to 2.5 ppm. If the tank holds 3.5 MGals, how many pounds of chlorine will be needed?

To determine how many pounds of chlorine are required to achieve a concentration of 2.5 parts per million (ppm) in a holding tank of 3.5 million gallons, we first need to understand the relationship between ppm, volume, and weight of the chemical.

1. Conversion of gallons to liters: Since 1 gallon is approximately 3.78541 liters, we can convert 3.5 million gallons into liters:

[

3.5 \text{ MGals} \times 3.78541 \frac{\text{liters}}{\text{gallon}} = 13,248,299 \text{ liters}

]

2. Understanding ppm: One ppm is equivalent to 1 mg of substance per liter of solution. Therefore, to achieve 2.5 ppm, you need 2.5 mg of chlorine per liter.

3. Calculating total mg of chlorine needed: To find the total amount of chlorine in milligrams for the entire volume, multiply the ppm concentration by the volume in liters:

[

2.5 \text{ mg/L} \times 13,248,299 \text{ L} = 33,120