1. Limiting Reagent
What happens in a chemical reaction, if there is an insufficient amount of one reactant?

2. In the laboratory, a reaction is rarely carried out with exactly the required amounts of each of the reactants. In most cases, one or more reactants is present in excess.
Once one of the reactants is used up, no more product can be formed.

3. Limiting Reagent: the reagent that is completely used up in a chemical reaction.
Excess Reagent: reagent not completely used up in a chemical reaction.

4. Example
C + O2
CO2
According to the balanced equation, 1 mole of carbon
reacts with 1 mole of oxygen to form 1 mole of carbon dioxide.
Suppose you mix 5 mole of carbon with 10 mole of oxygen
and allow the reaction to take place.
In this case there is more oxygen than is needed to
react with carbon. Carbon is the limiting reagent and limits the
amount of carbon dioxide formed.

5. Example: Find the limiting reagent when 1. 22g O2 reacts with 1
Example: Find the limiting reagent when 1.22g O2 reacts with 1.05g H2 to produce H2O.

6. Solution Use stoichiometry to calculate one of the products.
The reagent that gave the smaller calculated value of product is the limiting reagent.
The actual value of the amount of product formed is the smaller of the calculated values.

7. Example: Find the limiting reagent when 1. 22g O2 reacts with 1
Example: Find the limiting reagent when 1.22g O2 reacts with 1.05g H2 to produce H2O.
Use the grams to mole stoichiometry solution:
Calculate H2O moles produced by using each of the reactants:
Using O2: mol H2O.
Using H2: 0.5 mol H2O.
The actual amount H2O produced is the smaller one of the two values(0.076 mol H2O).
O2 is the limiting reagent, since O2 was used in the calculation of the mol H2O.

8. LIMITING REACTANT 4 Na (s) + O2(g)  2 Na2O (s)
a) Sodium metal reacts with oxygen to produce sodium oxide. If 5.00 g of sodium reacted with 5.00 grams of oxygen, how many grams of product is formed?
4 Na (s) + O2(g)  2 Na2O (s)
Start with what is given, calculate the amount of product that can be theoretically made but do it twice (once for each reactant):
5.00g Na (1 mole Na) ( 2 mole Na2O)( 62 g Na2O) = 6.74 g of Na2O
23 g Na mole Na mol Na2O
5.00g O2 (1 mole O2) ( 2 mole Na2O)( 62 g Na2O) = g of Na2O
32 g O mole O mol Na2O
Notice you can not have two different masses produced for the same product in one reaction vessel! So in this case, Na (sodium) “limits” how much sodium oxide is produced. The correct answer is 6.74 g of sodium oxide.
Wrong answer

9. LIMITING REACTANT 4 Na (s) + O2(g)  2 Na2O (s)
b) How much oxygen was used in this reaction and how much of each reactant was leftover (in excess)?
4 Na (s) + O2(g)  2 Na2O (s)
Use stoichiometry to compare the two reactants.
The amount of O2 used to make 6.74 g of Na2O is calculated by:
5.00g Na (1 mole Na) ( 1 mole O2)( 32 g O2) = 1.74 g of O2 was used
23 g Na mole Na mol O2
The amount of oxygen (O2) leftover can be calculated by subtracting the starting mass of oxygen from the used mass g – 1.74 g = 3.26 g in excess
The amount of sodium (Na) leftover at the end of the reaction is “0.00 g” (zero), since it was the limiting reactant and was completely consumed in the reaction.

10. 3 Pb+2 H3PO4  Pb3(PO4)2 (s) + 3 H2 (g)
LIMITING REACTANT
How many grams of solid are formed when 10.0 g of lead reacts with 10.0 g of phosphoric acid?
3 Pb+2 H3PO4  Pb3(PO4)2 (s) + 3 H2 (g)
Start with what is given but do it twice:
10.0g Pb (1 mole Pb) ( 1 mole Pb3(PO4)2)(811 g Pb3(PO4)2) = 13.1 g Pb3(PO4)2
207 g mole Pb mole Pb3(PO4)2
10.0g H3PO4 (1 mole H3PO4)( 1 mol Pb3(PO4)2)(811 g Pb3(PO4)2)=41.4 g Pb3(PO4)2
98 g mole H3PO4 1 mole Pb3(PO4)2
You can not have two different answers for one question so in this case lead “limits” how much lead(II) phosphate that can be produced. The correct answer is 13.1 g.
Wrong answer

11. PRACTICE PROBLEMS
1. How many grams of AgCl product will be produced if g of BaCl2 reacted with excess AgNO3?
2. How many moles of carbon dioxide could be produced from g of C2H2 and g of O2?
3. How many grams of CO2 can be produced by the reaction of 35.5 grams of C2H2 and 45.9 grams of O2?
4. In the reaction between CH4 and O2, if 25.0 g of CO2 are produced, what is the minimum amount of each reactant needed?
5. Cu AgNO3  Cu(NO3) Ag
When 10.0 g of copper was reacted with 60.0 g of silver nitrate solution, 30.0 g of silver was obtained. What is the percent yield of silver obtained? g
13.6 mol
50.5 g
9.09 g of CH4 & 36.4 g of O2
88.3%

12. PRACTICE PROBLEMS
______1. Which reactant will produce the least amount of AgCl product if reacted with AgNO3?
a) g BaCl b) g NaCl c) g CsCl
______2. How many moles of CO2 can be produced by the reaction of 5.0 grams of C2H4 and 12.0 grams of O2?
______3. How many grams of carbon dioxide could be produced from 2.0 g of C2H4 and 5.0 g of O2?
______4. In the reaction between CH4 and O2, if 18.0 g of CO2 are produced, how many grams of water are produced?
______5. Cu AgNO3  Cu(NO3) Ag.
When 50.0 g of copper was reacted with g of silver nitrate solution, 149 g of silver was obtained. What is the percent yield of silver obtained?