In a cyanide-bearing waste treatment process, how many pounds of chlorine are needed for the treatment of 6,000 gallons with a cyanide concentration of 15 mg/L?

To determine the amount of chlorine needed to treat 6,000 gallons of water containing cyanide at a concentration of 15 mg/L, we first convert the volume of water from gallons to liters, since chemical concentrations are typically expressed in milligrams per liter (mg/L).

1. Convert gallons to liters:

6,000 gallons is approximately 22,700 liters (using the conversion factor of 1 gallon = 3.78541 liters).

2. Calculate the total mass of cyanide in milligrams:

The concentration of cyanide is 15 mg/L, so for 22,700 liters, the total amount of cyanide would be:

[

\text{Total cyanide (mg)} = \text{concentration (mg/L)} \times \text{volume (L)} = 15 , mg/L \times 22,700 , L = 340,500 , mg

]

3. Convert milligrams to pounds:

There are approximately 453,592 mg in a pound. To convert the total cyanide mass to pounds, you use the following calculation:

[

\text{Total cyanide (pounds)} = \