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 8.5X10⁴ kg of coal that is 0.33 mass % sulfur?

Answer 1

To completely eliminate all of the SO2 produced by coal burning, 1250.62 kg of CaCO3 are required.

Equation

CaCO₃ --->CaO + CO2

2 CaO +2 SO2 ----> 2 CaS + 3O2

overall reaction

2 CaCO₃ +2 SO2 ---> 2 CaS + 3O2 + 2CO2

m(S)=8,5 x 10^{4}*0.0033=280.5 kg

n(SO2)= m/Ar(S) = 8,75 x 10^{3}mol

n(CaCO3 needed) = n(SO2) / percent yield

n(CaCO3 needed) = 8.75 x 10^{3}mol / 0.7

​n(CaCO3 needed)= 1.25 x 10 ^4 mol

m(CaCO3 needed) = n x Mr = 1250.62kg

To completely eliminate all of the SO2 produced by coal burning, 1250.62 kg of CaCO3 are required.

Where Is Limestone Found?

Numerous locations that produce limestone have been recognized by geologists throughout the world. Between 30o N and 30o S latitude, the most of them are in shallow waters. The Caribbean Sea, Indian Ocean, Persian Gulf, and the Gulf of Mexico are the most well-known shallow water locations.

learn more about limestone here

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