Theoretical Yield
The theoretical yield is the amount of product that would be produced in an ideal situation. Calculating the theoretical yield is easy.
We take the steps we have from finding limiting reagents, and add a few more steps to them.
Lets look at an example:
For the balanced equation shown below, if 37.7 grams of C6H6S2 were reacted with 68.9 grams of O2, how many grams of SO3 would be produced?
2C6H6S2+21O2=>12CO2+6H2O+4SO3
So the first step is finding the limiting reagent.
C = 12
H = 1
S = 32
O = 16
12 * 6 = 72
1 * 6 = 6
32 * 2 = 64
72 + 6 + 64 = 142 grams/mole
37.7/142 = .26 moles of C6H6S2
16 * 2 = 32 grams/mole
68.9/32 = 2.15 moles of O2
2 * .26
21 2.15
2 * 2.15 = 4.3 moles of O2.26 * 21 = 5.46 moles of C6H6S2
Limiting Reagent = O2
So, now that we know what the limiting reagent is, we need to look at the coefficient of the limiting reagent and the coefficient of the product in the chemical equation.
the ratio would be: 21:4
Now we need to multiply the molar mass of the limiting reagent by its coefficient and the molar mass of the product by its coefficient.
First lets find the molar mass of SO3
32 * 1 = 32
16 * 3 = 48
32 + 48 = 80 grams/mole
32 * 21 = 672
80 * 4 = 320
Now we divide the grams of the product by the grams of the limiting reagent.
320/672 = .476
Finally we take that answer and multiply it by the number of grams of the limiting reagent from the original question.
.476 * 68.9 = 32.8 grams
theoretical yield = 32.8 grams