J.R. S. answered 12/03/18
Ph.D. University Professor with 10+ years Tutoring Experience
a.) Your balanced equation looks correct.
b.) The heat for 1 mole CaCO3 (100. g) = 178.1 kJ. Thus, 178.1 kJ/100.0 g x 18.0 g = 32.1 kJ (3 sig. figs.)
Radi D.
asked 12/03/18The decomposition of limestone, CaCO3 into lime, CaO, and CO2 requires the addition of 178.1 kJ of heat per mole of CaCO3.
a. Write a balanced equation. (I balanced this to CaCO3 --> CaO + CO2)
b. How much heat is needed to decompose 18.0g of CaCO3
J.R. S. answered 12/03/18
Ph.D. University Professor with 10+ years Tutoring Experience
a.) Your balanced equation looks correct.
b.) The heat for 1 mole CaCO3 (100. g) = 178.1 kJ. Thus, 178.1 kJ/100.0 g x 18.0 g = 32.1 kJ (3 sig. figs.)
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