How many pounds of chlorine are required to treat a cyanide-bearing waste of 6,000 gallons at a concentration of 15 mg/L using alkaline chlorination?

To determine how many pounds of chlorine are required for the treatment of cyanide-bearing waste using alkaline chlorination, one needs to understand the chemical requirements for the process.

The first step is to convert the waste volume from gallons to liters since the concentration is given in mg/L. There are approximately 3.78541 liters in a gallon, so 6,000 gallons converts to about 22,700 liters (6,000 x 3.78541).

Next, using the concentration of cyanide (15 mg/L), we can find the total mass of cyanide in the waste. This is calculated by multiplying the volume in liters by the concentration in mg/L:

Total cyanide mass = Volume (liters) x Concentration (mg/L)

Total cyanide mass = 22,700 L x 15 mg/L = 340,500 mg.

To convert this mass to grams, we divide by 1,000 (as there are 1,000 mg in a gram):

Total cyanide mass in grams = 340,500 mg / 1,000 = 340.5 grams.

In alkaline chlorination, the stoichiometric requirement is typically around 1:1 for cyanide to chlorine. However