Excess & Limiting Reactants persent yield

A.Excess Quantity
A balanced equation describes what should happen in a chemical reaction.

However, the conditions necessary for the reaction to take place may not be present. (ie. pressure, temperature, concentration etc.)

Sometimes it is necessary to add more of one reactant than the equation predicts because it is not possible for every atom/molecule of the reactants to come together.
Ex. One reactant is the Excess Quantity and some of it will be left over, the second reactant is used up completely, and is the limiting reactant.

B.Excess quantities in chemical reactions.
Ex1. How many grams of OCl2 will be formed when 44.0 g of O2 reactant with 97.0g of Cl2

(*General Process: Convert both reactants to the desired product & the smaller amount of product will actually be produced.)

Step1: Balanced equation.
O2(g) +2 Cl2(g) → 2OCl2(g) 

Step2: Convert Cl2 to OCl2
97.0 g ×1mol Cl2/71g × 2mol OCl2/2mol Cl2 × 87.0g/1mol OCl = 118.86 g OCl2

Step3: Convert O2 to OCl2
44.0g × 1molO2/32g ×2mol OCl2/1mol O2 × 87.0g/1mol OCl = 239.25g OCl2
Because 97.0g of Cl2 reacts to produce the smaller amount of product. Cl2 is limiting quantity and O2 is excess quantity. 119g is the mass of product.

Ex2. 
41g O2 reacts of 164g of Cl2, Which is EXCESS quantity? Which is Limiting quantity? How many grams of the excess quantity will be used?
Step1:
O2(g) +2 Cl2(g) → 2OCl2(g) 
41.0g       164g
(*General process: Convert ONE of products to the other reactant to see which excess and which is limiting. Determine which is left over and by how much. It may involve another conversation)
Step2:
To find which reatant is in excess calculate how many grams of Cl2 gas would be required to react with 41.0g of O2.
41g O2 × 1mol O2/32.0gO2 × 2mol Cl2/1mol H2 × 71.0g Cl2/1mol Cl2= 181.93g=182g Cl2

Since there are only 164g of Cl2 gas, not all O2 can react. Thus Cl2 is limiting quantity and O2 is excess quatity.

Step3: Convert Cl2 to O2 to see how much O2 would be needed to react with 164g of Cl2.
164g Cl2 × 1mol Cl2/71.0g Cl2 × 1mol O2/2mol Cl2 × 32.0g/1mol O2 = 36.958=37.0 g O2

37.0g of O2 would react with 164g of Cl2.