I can just give you an example I guess
Just say you performed this reaction:
H2SO4 + 2NaOH -> 2H2O + Na2SO4
and you got a yield of 5.0 grams of sodium sulphate (Na2SO4)
and you reacted
10mL 18M H2SO4
and
20grams of NaOH
so first convert both to moles
To convert H2SO4 to moles we use the formula
Molarity = Moles/Liters
so
18M = x moles/0.01L
18M*0.01 = 0.18mol
Now convert NaOH to moles, using the formula I posted above
20g/(22.989g/mol + 15.9994g/mol + 1.00794g/mol)
=
0.5mol NaOH
so we have,
0.5moles of NaOH
and
0.18moles of H2SO4
ok now we figure out what the limiting reagent is by converting both to moles of the desired product, what ever gives us the least amount of moles product is our limiting reagent
Lets find out if NaOH is our limiting reagent
look at reaction again
H2SO4 + 2NaOH -> 2H2O + Na2SO4
we are going from NaOH to Na2SO4, note the coefficients, we multiply by 2 and divide by 1
( (0.5moles)(2) )/1 = 1mol of Na2SO4
Now lets find out if H2SO4 is our limiting reagent
H2SO4 + 2NaOH -> 2H2O + Na2SO4
Same deal, this time though we multiply by 1 and divide by 1, (this is a 1 to 1 reaction)
( (0.18moles)(1) )/1 = 0.18mol of Na2SO4
we see that the H2SO4 gives us the smallest amount of product thus it is the limiting reagent when it is used up the reaction ends (In this case it is an equilibrium reaction though but dont worry about that)
Thus we have a theoretical yield of 0.18mol of Product
Now we just need to find moles of product obtained
5.0 grams/(142.02g/mol) = 0.0352mol
so we got 0.0352mol of product and a theoretical yield of 0.18mol
now we just apply the formula I stated in my first post to find the percent yield
(0.0352mol/0.18mol)*100 = 19.6%