Question

Calculate the mass of lime that can be prepared by heating 200 kg of
limestone that is 90% pure CaCO3

Answer 1

mass of  CaCO₃ in 200 grams of 90% pure calcium carbonate is 180 grams
no. of moles of CaCO₃ in 180 grams of CaCO₃ is 180/100=1.8 moles

so, 1.8 moles each of CaO and CO₂ are formed by Decomposition of 1.8 moles of CaCO₃.

CaCO₃------------Δ------------->>> CaO + CO₂

mass of 1 mole CaO is 56 grams
mass of 1.8 moles of CaO is 56*1.8=100.8 grams

Answer 2

Answer:

Here is the answer......