Chemistry help

I don't understand it at all.

Solid calcium carbonate, CaCo3, is able to remove sulfur dioxide from waste gases by the reaction:

CaCO3 + SO2 + other reactants = CaCO3 + other products

In a particular experiment, 255 of CaCO3 was exposed to 135 g...

of SO2 in the presence of an excess amount of the other chemicals required for the reaction.

What is the theoretical yield of CaSO3?

If only 198 g og CaSO3 was isolated from the products, what was the percentage yield of CaSO3 in this experiment?

I don't understand how to find the theoretical yield or any of what this is saying!
Help!

What they want you to do is something called stoichiometry. Theoretical yield is how much product SHOULD be made, and we can find it by doing a mass to mass conversion.
Steps:
1. Make CERTAIN you have a balanced equation. If it's not balanced, all your calculations will be incorrect. (I'm assuming the equation is correctly balanced since they do not specify all of the reactants and products)
2. Change grams of the limiting reactant (or in this case, the element they want the theoretical yield for) to moles of reactant by dividing the number of grams by themolar mass of the compound. We see this below:

CaCO3= 100 g/mol
255 g CaCO3 / 100 g/mole= 2.55 mol CaCO3

As you can see, the grams divided with grams, leaving our answer in moles, which was our orignial goal.

3. Use mole ratios to determine how much product can be made. The formula is:
moles of known x moles of unknown/ moles of known =moles of unknown

*The first moles of known is the number of moles given in the problem. Moles of unknown will be the coefficient in front of the compound of unknown in the equation. The second moles of known will be the coefficient in front of the compound of known in the equation.

We set up the equation here:

2.55 mol CaCO3 X 1 mol CaSO3/ 1 mol CaCO3
= 2.55 mol CaSO3

The mol CaCO3s canceled out, leaving our answer in mol of CaSO3.

4. Change the moles of product obtained in the previous step back into grams by multiplying by molar mass.

Molar mass CaSO3 = 120 g/mole
120 g/mol X 2.55 mol = 306 g CaSO3
This is your theoretical yield.

To determine percent yield, divide actual yield by theoretical yield and multiply by 100.

Formula: (Actual Yield/ Theoretical Yield) X 100

(196 g / 306 g) x 100= 64.1 %

Hope this helps