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Answer to Question #73739 in Inorganic Chemistry for maddy
Question #73739
How many grams of oxygen are required to react with 32.5 g of lead(II) sulfide?
Expert's answer
2PbS + 3O2 -> 2PbO + 2SO2
(32.5 g PbS) / (239.265 g PbS/mol) x (3 mol O2 / 2 mol PbS) x (31.99886 g O2/mol) = 6.52 g O2
(32.5 g PbS) / (239.265 g PbS/mol) x (3 mol O2 / 2 mol PbS) x (31.99886 g O2/mol) = 6.52 g O2
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