Ca(OH)2 + SO2 ----> CaSO3 + H2O
mols SO2 = 1000/molar mass SO2 = approx 15.6 but you need to confirm that as well as all of the other numbers I do becuase I estimate some of them.
Convert to mols Ca(OH)2 and
15.6 mols SO2 x (1 mol Ca(OH)2/1 mol SO2) = 15.7.
Grams Ca(OH)2 = mols x molar mass = approx 15.7 x about 74 = about 1160 g if we are talking 100% efficient.
If only 80%, then it will take 1160/0.80 = ?
Ca(OH)2 + SO2 ----> CaSO3 + H2O
Assuming that this process is only 80 % efficient, how much Ca(OH)2 would be required to remove 1000 grams of SO2?
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