EXCESS & LIMITING REACTANT

Excess and Limiting reactant

2H₂ + O₂ 

→

 2H₂O

If 4 grams of hydrogen reacts with 16 grams of oxygen what mass of water is formed?
Step 1 Convert to mole all the reactants
Moles of hydrogen = 4grams/2 atomic mass units = 2mole
Moles of Oxygen gas = 16 grams/ 32 atomic mass units = 0.5mole

Step 2 Identify the reactant that is the LIMITING reactant. (In other words the reactant that will stop the reaction and the formation of product)
Take hydrogen and ask the following questions.How much do we have? = 2moles How much oxygen do we need to react with this amount of hydrogen? = 1/2 X 2moles = 1mole of oxygen.
 we have 0.5 mole of oxygen, therefore oxygen is our limiting reactant.

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + energy
 
What mass of carbon dioxide forms in the reaction of 25 grams of glucose with 40 grams of oxygen?
 
When approaching this problem, you should be able to see that for every 1 mole of glucose (C6H12O6), you need 6 moles of oxygen to obtain 6 moles of carbon dioxide and 6 moles of water.
 
Step 1: Determine the balanced chemical equation for the chemical reaction.
 
Already given!
 
Step 2: Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).
 
25g x (1 mol/180.06g) = 0.1388 mol  C6H12O6
 
40g x (1 mol/32g) = 1.25 mol O2                                                                                                                                               
 
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
 
a. If all of the 1.25 moles of oxygen were to be used up, there would need to be 1.25 x (1/6) or 0.208 moles of glucose. There is only 0.1388 moles of glucose available which makes it the limiting reactant.
 
     
 
1.25 mol O2 x (1 mol C6H12O6)/(6 mol O2) = 0.208 mol C6H12O6
 
b. If all of the 0.1388 moles of glucose were to be used up, there would need to be 0.1388 x 6 or 0.8328 moles of oxygen. Since there is an excess of oxygen, the glucose amount will be used to calculate the amount of the products in the reaction.
 
0.1388 mol C6H12O6 x (6 mol O2)/(1 mol C6H12O6) = 0.8328 mol O2
 
If more than 6 moles of O2 is available per mole of C6H12O6, the oxygen is in excess and the glucose is the limiting reactant. If less than 6 mole of oxygen is available per mole of glucose, the oxygen is the limiting reactant. Our ratio is 6 mole oxygen per 1 mole glucose, OR 1 mole oxygen per 1/6 mole glucose. This means: 6 mol O2 / 1 mol C6H12O6 . 
 

Therefore the mole ratio is: (0.8328 mol O2)/(0.208 mol C6H12O6)
 
Which gives us a 4.004 ratio of O2 to C6H12O6.
 
Step 4: Use the amount of limiting reactant to calculate the amount of CO2 or H2O produced.
 
For carbon dioxide produced: 0.1388 moles glucose x (6/1) = 0.8328 moles carbon dioxide.
 
Step 5: If necessary, calculate how much is left in excess.
 
1.25 mol - 0.8328 mol = 0.4172 moles of oxygen left over