Answer to Question #310421 in General Chemistry for JPDO

General Chemistry

Question #310421

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

Expert's answer

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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