What is Stoichiometry?
Stoichiometry comes from two Greek words, the first being Sthoicheion which means element, and the other is Metron which means measure. Stoichiometry is very mathematical. It involves calculating the masses, and sometimes volumes of reactants and products used in a chemical reaction.
4 Types of Stoichiometry Problems
1. Mole-Mole
Ex. How many moles of LiOH are required to react with 20 moles K2O?
Start with balanced equation.
Factor Label Method (fences).
One portion of fence will be a molar ratio from balanced equation.
2LiOH + K2O ---> Li2O + 2KOH
20 molesK2O | 2 moles LiOH = 40 moles LiOH
| 1 mole K2O
Stoichiometry comes from two Greek words, the first being Sthoicheion which means element, and the other is Metron which means measure. Stoichiometry is very mathematical. It involves calculating the masses, and sometimes volumes of reactants and products used in a chemical reaction.
4 Types of Stoichiometry Problems
1. Mole-Mole
Ex. How many moles of LiOH are required to react with 20 moles K2O?
Start with balanced equation.
Factor Label Method (fences).
One portion of fence will be a molar ratio from balanced equation.
2LiOH + K2O ---> Li2O + 2KOH
20 molesK2O | 2 moles LiOH = 40 moles LiOH
| 1 mole K2O
2. Mass-Mole
Ex. If 824 grams of NH3 and excess oxygen gas react to produce nitrogen oxide and water, how many moles of nitrogen oxide are formed?
2NH3 + 3O2 ---> N2O3 + 3H2O
824 gNH3 | 1 mole NH3 | 1 mole N2O3 = 24.2 moles N2O3
| 17.034 gNH3 | 2 moles NH3
Ex. If 824 grams of NH3 and excess oxygen gas react to produce nitrogen oxide and water, how many moles of nitrogen oxide are formed?
2NH3 + 3O2 ---> N2O3 + 3H2O
824 gNH3 | 1 mole NH3 | 1 mole N2O3 = 24.2 moles N2O3
| 17.034 gNH3 | 2 moles NH3
3. Mass-Mass
Ex. How many grams of LiOH are needed to react with 25.6 grams of Al2O3?
6LiOH + Al2O3 ---> 3Li2O + 2Al (OH)3
25.6 g Al2O3 | 1 mole Al2O3 | 6 moles LiOH | 24 g LiOH = 36.2 g LiOH
| 101.96 g Al2O3| 1 mole Al2O3 | 1 mole LiOH
Ex. How many grams of LiOH are needed to react with 25.6 grams of Al2O3?
6LiOH + Al2O3 ---> 3Li2O + 2Al (OH)3
25.6 g Al2O3 | 1 mole Al2O3 | 6 moles LiOH | 24 g LiOH = 36.2 g LiOH
| 101.96 g Al2O3| 1 mole Al2O3 | 1 mole LiOH
4. Volume-Volume (22.42 = 1 mole anything at STP, Standard Temperature, Pressure)
Ex. What volume of O2 is required for combustion of 0.35 L of C3H8?
C3H8 + 5O2 ---> 3CO2 + 4H2O
0.35 L C3H8 | 1 mole | 5 moles O2 | 22.42 L O2 = 1.75 L O2
| 22.42 L C3H8 | 1 mole C3H8 | 1 mole 2
Ex. What volume of O2 is required for combustion of 0.35 L of C3H8?
C3H8 + 5O2 ---> 3CO2 + 4H2O
0.35 L C3H8 | 1 mole | 5 moles O2 | 22.42 L O2 = 1.75 L O2
| 22.42 L C3H8 | 1 mole C3H8 | 1 mole 2
Limiting Reactants
Usually, in chemistry, you will have more of one reactant needed to completely use up the other reactant. That reactant is said to be excess, there is too much. The other reactant limits how much product we get, and once it runs out the reaction stops. This reactant is called limiting.
Ex. 10.0 grams of Aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is being produced?
2Al + 3Cl2 ---> 2AlCl3
10 g Al | 1 mole Al | 3 moles Cl2 | 70.90g Cl2 = 39.4 g Cl2 needed
| 26.98g Al | 2 moles Al | 1 mole Cl2 35.0 g Cl2 given
Cl- limiting
Al- excess
35.0g Cl2 | 1 mole Cl2 | 2 moles AlCl3 | 133.33g AlCl3 = 44.0 g AlCl3 is produced
| 70.9g Cl2 | 3 moles Cl2 | 1 mole AlCl3
Always use the limiting reaction when you are finding how much product is being produced
Usually, in chemistry, you will have more of one reactant needed to completely use up the other reactant. That reactant is said to be excess, there is too much. The other reactant limits how much product we get, and once it runs out the reaction stops. This reactant is called limiting.
Ex. 10.0 grams of Aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is being produced?
2Al + 3Cl2 ---> 2AlCl3
10 g Al | 1 mole Al | 3 moles Cl2 | 70.90g Cl2 = 39.4 g Cl2 needed
| 26.98g Al | 2 moles Al | 1 mole Cl2 35.0 g Cl2 given
Cl- limiting
Al- excess
35.0g Cl2 | 1 mole Cl2 | 2 moles AlCl3 | 133.33g AlCl3 = 44.0 g AlCl3 is produced
| 70.9g Cl2 | 3 moles Cl2 | 1 mole AlCl3
Always use the limiting reaction when you are finding how much product is being produced
Percent Yield
Percent yield is a ratio between actual yield and theoretical yield, shown as a percentage.
Theoretical Yield- Represents maximum amount of product that can be formed from the given amounts of reactants.
Actual Yield- Amount of product that actually forms when a reaction is carried out in a lab. Considered experimental value.
Percent Yield = Actual Yield x 100
Theoretical Yield
Ex. What is the percent yield is a student makes 5.2 grams (actual) of carbon dioxide by decomposing 9.5 grams of aluminum bicarbonate?
2 Al(HCO3)3 ---> 1 Al2O3 + 6 CO2 + 3 H2O
9.5 g Al(HCO3)3 | 1 mole Al(HCO3)3 | 6 moles CO2 | 44g CO2 = 5.97 g CO2
| 210.034g Al(HCO3)3 | 2 moles Al(HCO3)3 | 1 mole CO2 (Theoretical)
Percent yield is a ratio between actual yield and theoretical yield, shown as a percentage.
Theoretical Yield- Represents maximum amount of product that can be formed from the given amounts of reactants.
Actual Yield- Amount of product that actually forms when a reaction is carried out in a lab. Considered experimental value.
Percent Yield = Actual Yield x 100
Theoretical Yield
Ex. What is the percent yield is a student makes 5.2 grams (actual) of carbon dioxide by decomposing 9.5 grams of aluminum bicarbonate?
2 Al(HCO3)3 ---> 1 Al2O3 + 6 CO2 + 3 H2O
9.5 g Al(HCO3)3 | 1 mole Al(HCO3)3 | 6 moles CO2 | 44g CO2 = 5.97 g CO2
| 210.034g Al(HCO3)3 | 2 moles Al(HCO3)3 | 1 mole CO2 (Theoretical)
Percent Yield = 5.2 x100
5.97
Percent Yield = 87.1 %
5.97
Percent Yield = 87.1 %