To neutralize a sour digester, if 1 mg/L of lime needs to be added for every mg/L of volatile fatty acids present in the sludge with a concentration of 760 mg/L, how many pounds of lime should be added to 242,000 gallons of sludge?

To determine the amount of lime needed to neutralize the sour digester, the calculation starts with understanding the concentration of volatile fatty acids in the sludge. For every 1 mg/L of volatile fatty acids, an equal amount of lime (1 mg/L) is required for neutralization.

The given concentration of volatile fatty acids is 760 mg/L. Therefore, for 760 mg/L of volatile fatty acids, a corresponding 760 mg/L of lime is required.

Next, we need to convert the gallons of sludge into liters, since the concentration is expressed in mg/L. There are approximately 3.78541 liters in a gallon. Thus, 242,000 gallons can be converted as follows:

242,000 gallons * 3.78541 L/gallon = 915,620.42 liters.

To find out the total mass of lime needed in milligrams, we multiply the required amount of lime (760 mg/L) by the total volume in liters:

760 mg/L * 915,620.42 L = 694,379,116.8 mg.

Since the final answer is preferred in pounds, we convert milligrams to pounds, knowing that 1 pound equals 453,592.37 milligrams:

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