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Answer to Question #310421 in General Chemistry for JPDO

Question #310421

The tons of limestone(CaCO3) needed for the preparation of 5 tons of lime (CaO)



1
Expert's answer
2022-03-13T15:18:09-0400

CaCO3 = CaO + CO2

100 g/mol CaCO3 - 56 g/mol CaO

x tons CaCO3 - 5 tons CaO

x = 100*5/56 = 8.93 tons


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