For the balenced equation shown below, if 21.0 grams of C2H4 were reacted with 139 grams of O2, how many grams of H20 would be produced?
CH4+3O2 -> 2CO2+2H2O

first you find the limiting reagent


Reagent 1 = 21.0g of C2H4
Reagent 2 = 139g of O2
C2H4 = 28
O2 = 3232 x 3  x 21 = 72  = C2H4  is the limiting reagent
28 x 1


then you set up the equation

theoretical yield (moles) =    mass of L.R.      x   coefficent of product    
                                        molecular mass         coefficent of L.R. 


then you sub in the corresponding numbers


theoretical yield (moles) =   21  x  2  = 1.5
                                        28      1 
now that you have the theoretical yield in moles you convert it into grams by setting up this equation


Theoretical yield (grams) = (# of moles) x (molar mass of product)


then sub in the corresponding numbers


theoretical yield (grams) = (1.5) x (18) = 27


so the answer is 27 grams of H20