Limestone (CaCO3) is used to remove acidic pollutants from smokestack flue gases

Question

Limestone (CaCO3) is used to remove acidic pollutants from smokestack flue gases. It is heated to form lime (CaO), which reacts with sulfur dioxide to form calcium sulfite. Assuming a 70.% yield in the overall reaction, what mass of limestone is required to remove all the sulfur dioxide formed by the combustion of 7.2 104 kg of coal that is 0.33 mass % sulfur?

Explanation / Answer

sulphur mass =(0.33 x7.9 x10^4/100) = 260.7 kgs moles of S = 260700/32 = 8146.875 moles of SO2 = 8146.875 moles of CaO required = 8146.875 , since reaction is 1mole CaO+ 1SO2--> 1CaSO3 since yield is 70% moles of CaO required = (8146.875 x100/70) = 11638.4 moles of CaCO3 = 11638.4 (since 1CaCO3-->1CaO + 1CO2) mass of CaCO3 = 11638.4 x100 = 1163840 gm = 1163.84 kgs mass of limestone = 1163.84 kgs

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.