Ametallurgical firm needs to dispose 2.77 of gallons of waste sulfuric acid whose concentration has been determined to be 2.10 m. environmental regulations require them to neutralize the waste before releasing the wastewater, and they choose to do so with slaked lime, . the neutralization reaction is if the slaked lime costs $0.53 per pound, how much will it cost to neutralize the waste sulfuric acid?

It will cost 1.9$ to neutralize the waste sulfuric acid

Explanation:

Step 1: The neutralization reaction

H2SO4(aq) + Ca(OH)2 (s) → CaSO4(s) + 2H2O

Step 2: Calculate the molar mass of Ca(OH)2

Molar mass of Ca = 40.08 g/mole

Molar mass of O = 16 g/mole

Molar mass of H = 1.01 g/mole

Molar mass of Ca(OH)2 = 40.08 + 2*16 + 2*1.01 =74.1 g/mole

Step 3: Convert gallons to liters

2.77 gallons of H2SO4 = 10.48559 Liters H2SO4

Step 4: Calculate number of moles of H2SO4

Molarity = moles / liters

2.10 M = moles / 10.48559L

number of moles = 2.10 * 10.48559 =22.02 moles

Step 5: Calculate the number of moles of Ca(OH)2

H2SO4(aq) + Ca(OH)2 (s) → CaSO4(s) + 2H2O

For 1 mole of H2SO4 consumed we need 1 mole of Ca(OH)2

This means we need 22.02 moles of Ca(OH)2

Step 6: Calculate mass of Ca(OH)2

mass of Ca(OH)2 =number of moles * Molar mass of Ca(OH)2

mass of Ca(OH)2 = 22.02 moles * 74.1 g/moles = 1631.682 grams

Step 7: Convert grams to pounds

1631.682 grams/ 453.592 = 3.5972 pounds

Step 8: Calculate the cost

price Price per pound = 0.53$

Price for 3.5972 pounds = 0.53 * 3.5972 = 1.906516 ≈1.9$

It will cost 1.9$ to neutralize the waste sulfuric acid

po-215 i got it right on the quiz