mass coal is 8.5E4 kg or 8.5E7 g.
That is 0.33% S or 8.5E7 x 0.0033 = about 300,000 g but that's just an estimate. You should redo this and all calculations that follow to gt a more accurate answer. So how much SO2 is formed?
S + O2 ==> SO2
mols S = g/atomic mass = about 300,000/32 = about 10,000
Convert that to mols CaCO3.
CaCO3 ==> CaO
CaO + SO2 ==> CaSO3
10,000 mols S will use 10,000 mols
CaO which must have had 10,000 mols CaCO3
g CaCO3 = mols CaCO3 x molar mass CaCO3 = about 1,000,000 g or about 1,000 kg limestone. Post your work if you get stuck.
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 required to remove all the sulfur dioxide formed by the combustion of 8.5x10^4 kg of coal that is 0.33 mass% sulfur?
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