Answer: Detailed problem with lots of assumptions. Please see below. Percent yield is 100% (!?, so check my work).
Explanation:
Some information is missing from the problem statement. I'll make some assumptions to solve the problem, so this answer depends on those assumptions.
"Calcium carbobnate" is CaCO3, not CaCO, which is obtained by calcining (heating) CaCO3:
CaCO3 = CaO + CO2
The CaO can then be reacted with SO2 to form calcium sulfate, CaSO4 (not CaSO, as per the following balanced equation:
2CaO + 2SO2 + O2 = CaSO4
[Note: It also reacts with any SO3 molecules bouncing around, as does happen in flue gases to also form CaSO4]
The following assumes we have 255 grams of CaCO3, not CAC, whatever that stands for.
We have 135g of SO2, not SO.
This also assumes that it is CaSO4 that is produced (not CaSO), 198 grams.
Since is a two step reaction sequence, we need to first calculate the moles of Ca) produced from the CaCO3. It is a 1:1 molar ratio, and we have (255g/100.1g/mole) = 2.548 moles of CaCO3. This will produce 2.548 moles of CaO, for use in the second reaction.
2CaO + 2SO2 + O2 = 2CaSO4
This tells us we'll produce 1 mole of CaSO4 for every 2 moles of CaO, a 1:2 ratio. It also states that 2 moles of SO2 are consumed for every 2 moles CaO, a 1:1 ratio.
We have 2.548 moles of CaO, so would need the same moles of SO2 to fully react. However, we only have (135g/64.1 g/mole) = 2.107 moles of SO2. The SO2 will be the limiting reagent, so the theorectical yield will be based on this reactant, which will consume 2.107 moles of CaO, leaving 0.44 moles (24.7 g) of CaO behind.
The balanced equation tells us that 2.107 moles of CaO should produce 2.107 moles of CaSO4, which is (2.107moles)*(136.1 g/mole) = 198 grams of CaSO4. We actually got 198 grams, so the yield is 100% [Nice work].
