Hello, thank you for taking the time to post your question!
Whenever you are completing a stoichiometry problem like this you want to start by putting together the balanced chemical equation. In this case we have
CaCO3 --> CaO + O2 for the decomposition of limestone
From there then you want to take 100 g x 90% = 90g of CaCO3 based on the purity percentage
Convert that over to moles using the molar mass: 90g / (100g/mol) = 0.9 mol CaCO3
Then you can convert that over into moles of the other substance based on the 1:1 ratio in the equation … 0.9 mol x 1mol CaCO3 / 1mol CaO = 0.9 mol CaO
The final step then is to convert that back into grams by multiplying by the molar mass
0.9 mol x 56 g/mol = 50.4 grams of CaO
Hopefully that gets you moving in the right direction! Feel free to reach out if you have any questions beyond that on the simplification :)
