nCaCO3 = \(\dfrac{100}{100}=1\) mol
Pt: CaCO3 --to--> CaO + CO2
1 mol-------------> 1 mol
mCaO thu được = 1 . 56 = 56 (g)
c) nCaO = \(\dfrac{20}{56}\) mol
Pt: CaCO3 --to--> CaO + CO2
\(\dfrac{20}{56}\) mol<---------\(\dfrac{20}{56}\) mol
mCaCO3 cần dùng = \(\dfrac{20}{56}.100=35,71\left(g\right)\)