Equations and Stoichiometry

Equations and Stoichiometry I. Balancing Reactions II. Types of Reactions A. Synthesis 1. simple 2. acid anhydrides 3. basic anhydrides 4. amphoteric anhydrides B. Decomposition C. Single Replacement D. Double Replacement E. Neutralization F. Combustion III. Stoichiometry A. Mole to mole B. Liter to liter C. Gram to gram IV. Limiting Reagents V. % Yield Equations and Stoichiometry I. Reactions- General

Law of Conservation of Mass - Matter can not be created or destroyed in a chemical rxn, only changed.

MASSreactants = MASSproducts Law of Conservation of Energy - Energy can not be created or destroyed in a chemical rxn, only changed.

ENERGYreactants = ENERGYproducts

II. Balancing Reactions: # Moles (reactants) = # Moles (products) A. Steps to Solve: 1. Start with the hardest part 2. Even / Odd (make the odd even by x2) 3. Keep PAI together if possible 4. When done remember blanks are really 1 coefficients

B. Examples:

1. ___ H2 + ___ O2 --> ___ H2O 5

2. ___ N2 + ___ H2 --> ___ NH3 (Haber Process) 6

3. ___ C + ___ SO2 --> ___ CS2 + ___ CO 12

4. ___ CO2 + ___ H2O --> ___ C6H12O6 + ___ O2 19

5. ___ AgNO3 + ___ H2S --> ____ Ag2S + ___ HNO3 6

6. ___ Zn(OH)2 + ___ H3PO4 --> ___ Zn3(PO4)2 + ___ H2O 12 7. hydrogen + sulfur --> hydrogen sulfide 3 8. iron III chloride + calcium hydroxide --> iron III hydroxide + calcium chloride 10 C. Tougher Problems

1. ___ C7H6O2 + ___ O2 --> ___CO2 + H2O

2. ___ Na2O + ___ (NH4)2SO4 --> ___ Na2SO4 + ___ H2O + ___ NH3

3. ___ NH3 + ___ O2  ___ NO + ___H2O

D. Additional Practice Online Here Or Here If you really need help click here! When you've got it try the WORLD's MOST DIFFICULT BALANCING PROBLEM

III. Types of Reactions: VIDEO A. Synthesis (composition): •two or more elements or compounds may combine to form a more complex compound. •Basic form: A + B --> AB

1. General Examples: a. 2H2 + O2 --> 2H2O This is an important reaction used in fuel cells and to power the space shuttle. b. 2Na + Cl2 --> 2NaCl c. 8Fe + S8 --> 8FeS d. 2K(s) + Cl2(g) --> 2KCl(s)

2. Acid Anhydrides (nonmetal /molecular oxide) + H2O produce acids a. SO3 + H2O --> H2SO4 (oil of vitriol) b. NO2 + H2O --> H2NO3 c. CO2 + H2O --> H2CO3

3. Basic Anhydrides (metal oxide) + H2O produce bases a. Na2O + H2O --> 2NaOH (lye) b. MgO + H2O --> Mg(OH)2 (milk of magnesia)

4. Amphoteric Anhydrides (oxides that can act as either an acid or a base) These are typically composed of metalloid oxides. 5. Basic and Acidic Anhydrides combine to form salts a. Na2O + CO2 --> Na2CO3 b. CaO2 + SO2 --> CaSO4

B. Decomposition: •A single compound breaks down into its component parts or simpler compounds. •Basic form: AB --> A + B (These are the reverse of SYNTHESIS)

The following is a specific list of decomposition reactions. Just remember that reactions do not like to gain energy typically. So if a low energy molecule can be "squeezed out" of a compound it will be. Typically a simple salt or metal oxide is also left.

A LIST OF LOW ENERGY MOLECULES: CO2, H2O, N2, O2, SO3, NO2

1. General Examples: a. 2 FeCl3 (s) + energy --> 2 Fe(s) + 3 Cl2(g) b. 2 NaCl (aq) + electricity --> 2Na(s) + Cl2(g) c. CuCl2 (aq) + electricity --> Cu(s) + Cl2(g) d. 2 H2O + electricity --> 2 H2(g) + O2(g)

2. Acids - acids decompose with heat to produce an acid anhydride and H2O a. H2CO3 --> CO2 (g) + H2O (l) b. H2SO4 --> SO3(g) + H2O c. HNO3 + HNO2 --> 2 NO2 + H2O

- 3. Hydroxides "Bases" - m(OH ) decompose to form a basic anhydride and H2O a. NaOH (s) --> Na2O (s) + H2O b. Mg(OH)2 (s) --> MgO (s)+ H2O

4. Carbonates - decompose to form an oxide and liberate CO2 gas a. CaCO3(s) --> CaO and CO2 b. K2CO3 (s) --> K2O + CO2

5. Chlorates - decompose to form a metal chloride and liberate O2 gas a. 2KClO3 (s) + heat --> 2KCl + 3O2 *demo

6. Sulfates - decompose to form a metal oxide and liberate SO3 gas a. Na2SO4 (s) + heat --> Na2O + SO3(g)

C. Single Replacement: • a more active element takes the place of another element in a compound and sets the less active one free. Refer to Table J for reactivity. • Basic form: A + BX --> AX + B or AX + Y --> AY + X

1. Mg + HCl --> MgCl2 + H2 *demo

2. Na + H2O --> NaOH + H2

3. Zn(s) + CuSO4(aq) --> ZnSO4(aq) + Cu(s) *demo

4. K(s) + AgNO3(aq) --> KNO3(aq) + Ag

5. Cl2 + NaBr --> NaCl + Br2 D. Double Replacement •occurs between ions in aqueous solution. A reaction will occur when a pair of ions come together to produce at least one of the following: 1. a gas 2. a precipitate 3. water or some other low energy molecule. • Basic form: AX + BY --> AY + BX

1. (NH4)2S(aq) + MgCl2(aq) --> MgS(s) + NH4Cl(aq)

2. Pb(NO3)2(aq) + KI(aq) --> PbI2(s) + KNO3(aq) *demo

3. FeS(s) + 2HCl --> FeCl2(aq) + H2S(g) *demo

4. AgNO3(aq) + NaCl(aq) --> AgCl(s) + NaNO3(aq)

5. NaHCO3(s) + HC2H3O2(aq) --> NaC2H3O2 + H2O + CO2

E. Neutralization This is a special form of Double Replacement reaction that occurs between an acid(H/nm) and a base(mOH).

1. 2HNO3 + Mg(OH)2 --> Mg(NO3)2 + 2H2O

2. H2SO4 + 2NaOH --> Na2SO4 + H2O

3. HC2H3O2 + KOH --> KC2H3O2 + H2O

4. *NaHCO3(s) + HC2H3O2 --> NaC2H3O2 + H2O + CO2 (TWO TYPES OF REACTIONS HERE - neutralization and decomposition)

F. Combustion The combination of a fuel with an oxidizing agent (typically O or F) to release energy. These often involve organic molecules which produce CO2 and H2O along with energy.

1. CH4 + 2 O2 --> CO2 + 2 H2O

2. 2C2H2 + 5O2 --> 4CO2 + 2H2O 3. more practice IV. Stoichiometry (pg. 261-271 text) Now that you can write formulas, predict reactions and balance equations it is now possible to determine quantities of reactants and products that will be involved in every reaction. *Remember- the only reactions that can be mathematically determined directly are mol to mol and liter to liter.

A. Mole to mole problems - Given: moles Asked for: moles Steps to Solve: Set up a RATIO using coefficients.

N2(g) + 3 H2(g) --> 2 NH3(g)

1. How many moles of H2 are needed to produce 12 moles of NH3? Answer:

2. How many moles of N2 are needed to completely react with 1.5 moles of H2? Answer: X = .5 moles

3. How many moles of N2 must react to produce 10 moles of NH3? Answer: X = 5 moles

2 H2(g) + O2(g) --> 2 H2O(g)

B. Liter to Liter Problems - Given: liters Asked for: liters Steps to Solve: Set up a RATIO using coefficients

1. How many liters of O2 will react completely with 50 L of H2? Answer:

2. 2 Liters of O2 will react to form how many liters of H2O? How many liters of H2 will be needed? Answer: 4 L H2O , 4 L H2

C. Mole (<-->) Liter Problems - Steps to Solve: Set up a RATIO using coefficients, incorporate 22.4 with L!

1. How many liters of chlorine gas will be produced if 4 moles NaCl is decomposed? 2 NaCl(s) --> Na(s) + Cl2(g) 44.8L 2. How many moles of water will be produced if 18 liters of hydrogen react with oxygen to form H2O? 2 H2(g) + O2(g) --> 2H2O(g) 0.81 moles

D. Gram to Gram : Steps to Solve: Set up a RATIO using coefficients, include molecular weights of reactant and products.

2K + MgBr2 ---> 2KBr + Mg

1. How many grams of KBr will be produced if 100 grams of MgBr2 react completely? Step 1: KBr (119g/mol) , MgBr2 (181.4g/mol) Step 2: 100g/1(181.4g/mol) = x/2(119g/mol) Step 3: x = 130.9g KBr

E. Grams (<-->) Liters: Steps to Solve: Set up a RATIO using coefficients, include molecular weight of reactant and 22.4 for gas product.

F. Tougher Word Problems - click HERE

G. Limiting Reagents- Unless reactants are carefully measured out there will typically be one in excess. When these react, it is only possible to make as much product as the limiting reactant(reagent) will allow, according to the mole ratio in the balanced equation.

1. Limiting Reactant - The reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed.

2. Excess Reactant - The reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed. The excess reactant remains because there is nothing with which it can react.

No matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made. Likewise with chemistry, if there is only a certain amount of one reactant available for a reaction, the reaction must stop when that reactant is consumed whether or not the other reactant has been used up.

3. Steps to Solve: 1) Set up a RATIO using coefficients, include molecular weights of reactant and products. 2) Divide (lowest number is limiting reactant). 3) Set up Ratios of remaining members of the reaction and solve.

4. Examples: a. If 100 g K2S react with 150 g MgSO4 according to the following reaction:

K2S + MgSO4 --> MgS + K2SO4 (1) What is the limiting reactant?

Step 1: 100g K2S / 110.2 g/mol = 0.907 mol K2S

150g MgSO4 / 120.3g/mol = 1.25 mol MgSO4

Step 2: Divide.

.907 mol K2S 1.25 mol MgSO4 .907 is smaller so K2S is our limiting reactant.

(a) How many grams of magnesium sulfate will react and how much of each product will also be formed? K2S + MgSO4 --> MgS + K2SO4 .907 mol .907mol .907 mol .907 mol

100g 109.11g 51.06g 158.00g

MgSO4 - 0.907 mole (120.3 g/mol) = 109.11g MgSO4 MgS - .907 mol (56.3 g/mol) = 51.06g MgS

K2SO4 - .907 mol (174.2g/mol) = 158.00g K2SO4

Lets check our answer using the Law of Conservation of Mass:

100g K2S + 109.11g MgSO4 = 51.06g MgS + 158.00g K2SO4

209.11g reactants = 209.06g products considering rounding, we have conserved mass. b. More practice problems click HERE c. Homework Assignment pg 280. #9-14.

V. % Yield - describes the actual yield in an experiment compared to the theoretical yield expected.

A. Example: 1. What is the % yield if 40grams of a substance was produced, but 50 grams was expected?

2. What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2? answer: 95.8%

3. What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2? answer: 17.8% see work below