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Chemical Reactions Chapter 11

Chemical Reactions Chapter 11. SC2 Students will relate how the Law of Conservation of Matter is used to determine chemical composition in compounds and chemical reactions. SC2.a. Identify and balance the following types of chemical equations:

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Chemical Reactions Chapter 11

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  1. Chemical Reactions Chapter 11 SC2 Students will relate how the Law of Conservation of Matter is used to determine chemical composition in compounds and chemical reactions. SC2.a. Identify and balance the following types of chemical equations: Synthesis, Decomposition, Single Replacement, Double Replacement, and Combustion SC2.b. Experimentally determine indicators of a chemical reaction specifically precipitation, gas evolution, water production, and changes in energy to the system.

  2. Writing chemical Equations • In chemical reactions one or more reactants change into one or more products. • Chemists use a chemical equations to convey a s much information as possible about what happens in a chemical reaction. • The reactants are written on the left and the products on the right • An arrow separates them, read the arrow as yields, gives, or reacts to produce. Reactants → products • example: iron + oxygen → iron (III) oxide

  3. Writing chemical Equations • To write word equations, write the names of the reactants to the left of the arrow separated by plus signs; write the names of the products to the right of the arrow, also separated by plus signs. • Example: hydrogen peroxide decomposes to form water and oxygen gas • Hydrogen peroxide → water + oxygen • Chemical equation is a representation of a chemical reaction • Skeleton equation is a chemical equation that does not indicate the relative amounts of the reactants and products. • Write the formulas of the reactants to the left of the yields sign (arrow) and the formulas of the products to the right.

  4. Writing chemical Equations • You can indicate the physical states of substances by putting a symbol after each formula: (s) = solid, (l) = liquid, (g) = gas, and (aq) = aqueous solution [a substance dissolved in water] • Example: Fe (s) + O2 (g) → Fe2O3 (s) • Catalyst is a substance that speeds up the reaction but is not used up in the reaction. • A catalyst is neither a reactant nor product and is written above the arrow.

  5. Types of equations • Word equation • Carbon Dioxide + Water  Glucose + Oxygen • Skeletal equation • CO2 + H2O  C6H12O6 + O2 • Skeletal equation with phases • CO2 (g) + H2O (l)  C6H12O6 (s) + O2(g) • Balanced equation • 6CO2 + 6H2O  C6H12O6 + 6O2

  6. Types of Reactions • Combination (synthesis) reaction is a chemical change in which two or more substance react for form a SINGLE NEW SUBSTANCE. • Example: 2H2 + O2 → 2H2O • Combination reactions are easy to recognize because they have two reactants and one product. • Decomposition reaction is a chemical change in which ONE SUBSTANCE reacts to form TWO or more new substance. • Example: 2H2O → 2H2 + O2 • Decomposition reactions are easy to recognize because they have one reactants and two or more products.

  7. Types of Reactions 3. Single replacement reaction is a chemical change in which ONE ELEMENT replaces another element in a COMPOUND. • you can recognize a single replacement reaction because both the reactant and the product have an element and a compound • example: 2K + 2H2O → 2KOH + H2 4. Double-replacement reaction is a chemical change involving an exchange of positive ions between two compounds. • you can recognize a double-replacement reaction because both the reactants and the products are two compounds • They generally take place in aqueous solutions, and often produce a precipitate, a gas, or a molecular compound such as water. • FeCl3 + NaOH → Fe(OH)3 + NaCl

  8. Types of Reactions 5. Neutralization Reaction is a special type of double-replacement reaction. It is a chemical change where an acid and a base react to form water and a “salt” • Most bases are hydroxides, example: sodium hydroxide {NaOH} • Most acids start with hydrogen, example: hydrochloric acid {HCl}, sulfuric acid {H2SO4} • Salts are defined as ionic compounds • Example: NaOH + HCl → H20 + NaCl 6. Combustion reaction is a chemical change in which an element or a compound reacts with oxygen often producing energy in the form of heat and light. • Often the other reactant is a hydrocarbon or an alcohol and the products are carbon dioxide and water. • Example: CH4 + O2 → CO2 + H2O

  9. Types of Reactions Summary

  10. Types of Reactions Practice • Single Replacement • Double Replacement • Combination • Combination • Neutralization • Combustion • Decomposition • Single Replacement • Double replacement • Decomposition • Al + CH3OH  (CH3O)3Al + H2 • Ba(NO3)2+ CuSO4  BaSO4+ Cu(NO3)2 • N2 + H2  NH3 • H2O (l) + SO3 (s)  H2SO4(aq) • KOH + H3PO4  K3PO4 + H2O • C4H10 + O2  CO2 + H2O • H2O2 H2O + O2 • KI + Cl2KCl + I2 • AgNO3 + KClAgCl(s) + KNO3 • NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)

  11. Balancing chemical Reaction • The law of conservation of mass states that matter is not created or destroyed it merely changes forms. • In order to show that mass is conserved the amount of products and reactants MUST equal. • Once a word equation is turned into chemical formulas it must be balanced.

  12. Balancing Chemical Equations • Coefficients are small whole numbers that are placed in front of the formulas in an equation in order to balance it. • Balanced equation is a chemical equation in which each side of the reaction has the same number of atoms of each element and mass is conserved. • To write a balanced chemical equation, first write the skeleton equation. Then use coefficients to balance the equation so that it obeys the law of conservation of mass. • In every balanced chemical equation each side of the equation has the same number of atoms of each element.

  13. Steps to balancing chemical equations • Create a T-chart and List the elements in the chemical formulas on each side in the SAME order • Count the number of atoms of each element on each side • See what elements don’t match in your T-chart • Add coefficients in front of one substance at a time, • Start by balancing the most complex compound first, • general rule of thumb is to save Hydrogen and Oxygen until the end to balance. • recount number of atoms of each element • Repeat 3, 4 & 5 until all elements are balanced.

  14. Counting Atoms • In order to balance chemical reactions you MUST first know how to count atoms. • Subscripts multiply what ever they are beside • If a subscript is beside parentheses it multiplies everything in the parentheses. • Coefficients multiply everything immediately behind them.

  15. Counting Atoms Practice • 2 H, 1 O • 1 Cu ,1 S,4 O • 1 Na, 1 H, 1 C, 3 O • 1 Cu,2N,6 O • 3 N, 12 H, 1 P, 4 O • 4 H, 2 O • 3 Cu, 3 S, 12 O • 2 Cu, 4 N, 12 O • 9 N, 36 H, 3 P, 12 O • H2O • CuSO4 • NaHCO3 • Cu(NO3)2 • (NH4)3PO4 • 2 H2O • 3CuSO4 • 2 Cu(NO3)2 • 3 (NH4)3PO4

  16. Balancing Chemical Reactions Practice 3 2 Example 1: N2 + H2  NH3 N | N H | H • Pick an element to start balancing with, I pick N • Least common multiple between 1 and 2 is 2 • Add 2 in front of product side and recount • N is balanced so now look at H • Least common multiple between 2 and 6 is 6 • Add 3 in front of reactant H and recount • Now both sides are equal so it is a balanced equation = 1 = 3 = 2 = 6 = 2 = 6 = 2 = 2

  17. Balancing Chemical Reactions Practice • Is a trial and error process Example 1: NaHCO3  Na2CO3+ H2O + CO2 Na | Na H | H C | C O | O • Count Atoms of each element • Start balancing the most complex item (reactant) 2 Na = 1 | Na = 2 H = 1 | H = 2 C = 1 | C = 1+1 =2 O = 3 | O = 3 + 1 + 2 = 6 Na = 2 | Na = 2 H = 2 | H = 2 C = 2 | C = 1+1 =2 O = 6 | O = 3 + 1 + 2 = 6

  18. Balancing Chemical Reactions Practice 8 • Is a trial and error process Example 1: C4H10 + O2  CO2+ H2O C | C H | H O | O • Start balancing the most complex item (reactant) • Leave H and O until the end • Have to use whole number and currently have no whole number… trick: Double all current coefficients and recount. It is Now Balanced 2 13 4 10 5 C = 4 | C = 1 H = 10 | H = 2 O = 2 | O = 2 +1 = 3 C = 4 | C = 4 H = 10 | H = 2 O = 2 | O = 8 +1 = 9 C = 8 | C = 8 H = 20 | H = 20 O = 26 | O = 16 +10 = 26 C = 8 | C = 8 H = 20 | H = 20 O = 2 | O = 16 +10 = 26 C = 4 | C = 4 H = 10 | H = 10 O = 2 | O = 8 +5 = 13

  19. Balancing Reaction Practice – Polyatomic • Balancing reactions with polyatomics can be a bit complicated but if the polyatomic is the same on BOTH sides of reaction you can treat it like a special element. FeBr3 + Ba(OH)2 → BaBr2 + Fe(OH)3 • When counting Atoms put a box around JUST the polyatomic • Count like normal and balance using these count • Br is 1st thing that don’t balance, common multiple between 2 and 3 is 6 2 3 3 2 Fe 1 | Fe 1 Br 3 | Br 2 Ba 1 | Ba 1 (OH) 2 | (OH) 3 Fe 1 2 | Fe 1 Br 3 6 | Br 2 6 Ba 1 | Ba 1 3 (OH) 2 | (OH) 3 Fe 1 2 | Fe 1 2 Br 3 6 | Br 2 6 Ba1 3 | Ba 1 3 (OH)2 6 | (OH) 3 6 Fe 1 2 | Fe 1 2 Br 3 6 | Br 2 6 Ba 1 | Ba 1 3 (OH) 2 | (OH) 3 6

  20. Balancing Reaction Practice – Polyatomic 2 Na3PO4 + MgSO4 → Mg3(PO4)2 + Na2SO4 3 3 Na 3 6 | Na 2 6 PO41 2 | PO4 2 Mg 1 3 | Mg 3 SO41 3 | SO41 3 Na 3 6 | Na 2 6 PO41 2 | PO4 2 Mg 1 | Mg 3 SO4 1 | SO41 3 Na 3 | Na 2 PO4 1 | PO4 2 Mg 1 | Mg 3 SO4 1 | SO4 1 Na3PO4 + MgSO4 → Mg3(PO4)2 + Na2SO4 • When counting Atoms put a box around JUST the polyatomic • Count like normal and balance using these count • Na is 1st thing that don’t balance, common multiple between 2 and 3 is 6. • Fix each side and recount • Mg is next thing that don’t balance, common multiple between 1 and 3 is 3. Na 3 6 | Na 2 PO41 2 | PO4 2 Mg 1 | Mg 3 SO4 1 | SO4 1

  21. Predicting Products • Predicting the products of a chemical reaction involve 1st determining the type of reaction that is occurring. • Combination: starts with 2 elements • Decomposition: starts with 1 compound • Single Replacement: Starts with 1 element & 1 compound • Double Replacement: starts with 2 compounds • Neutralization: starts with 2 compounds that are an ACID and a BASE • Combustion: starts with O2 and a compound containing C, H and/or O.

  22. Predicting Products identifying types of reaction practice. Combination Neutralization Single replacement Combustion Decomposition Double replacement ____Al + ____Cl2  ___Mg(OH)2 + ___HCl ___Zn + ____HCl ___C2H6 + ___O2 ___Al2O3  __Ba(NO3)2 + __Na2SO4 

  23. Predicting Products • Predicting the products of a chemical reaction involve next determine the products that would be produced based on a balance of charge. • Combination: ends with 1 compound • Decomposition: ends with 2 elements • Single Replacement: ends with 1 element & 1 compound • Double Replacement: ends with 2 compounds • Neutralization: ends with H2O and a “salt” • Combustion: ends with CO2 and H2O

  24. Predicting Products • ____Al + ____Cl2  • Is a combination reaction so the product is a compound between aluminum and chlorine. MUST BALANCE CHARGES! • Al+3 and Cl-1 form compound • AlCl3 • ____Al + ____Cl2  ____ AlCl3still needs number of atoms to be balanced • ___Mg(OH)2 + ___ HCl • Is a neutralization reaction so the product is water and a salt. The salt would be formed between the element at the beginning of the base and the element at the end of the acid. Think square dance!! MUST BALANCE CHARGES! • Mg +2 and Cl-1 form compound • MgCl2 and H2O • ___Mg(OH)2+ ___HCl __ MgCl2 + __ H2Ostill needs number of atoms to be balanced

  25. Predicting Products • __Al2O3  • Is a decomposition reaction so the product is just elements. • Don’t forget diatomic elements. • Al and O2 (is a diatomic element) • __Al2O3  __ Al + ___ O2 • __Ba(NO3)2 + __Na2SO4  • Is a double replacement reaction, the positive ions switch partners. MUST BALANCE CHARGES! • __Ba(NO3)2 + __Na2SO4  __BaSO4 + __NaNO3 still needs to be balanced

  26. Predicting Products • ___Zn + ____HCl • Is a single replacement reaction so the element replaces an element in the compound • Generally the 1st element in compound is the one that leaves. MUST BALANCE CHARGES! • ___Zn + ____ HCl ___ZnCl2 + ___H2 • ___C2H6 + ___O2 • Is a combustion reaction, and can only make carbon dioxide and water. • ___C2H6 + ___O2 ___ CO2 + _____H2O

  27. Predicting Products Summary • Make sure to BALANCE charges for ALL ionic and acid compounds • Transition Metals use the reactant to determine the charge and assume that it stays the same. • Once you BALANCE CHARGES do NOT change the subscripts. • Don’t forget your 7 diatomic elements: • When they are not in compounds they are in pairs • N2, O2, F2, Cl2, Br2, I2, H2 • ALL other elements are single elements when not in compounds • Just because an element has a subscript on the reactant side, does NOT mean it will have the same subscript on the product side.

  28. Word Equations • When balancing reactions that are word equations you MUST first write BALANCED chemical formulas. • Remember to check the ions that each forms for Ionic compounds and Acids • Once you have a balanced chemical formula DO NOT change the formula (i.e. change subscripts) • DON’T forget your 7 diatomic elements.

  29. Word Equations • Example 1: solid calcium hydroxide reacts with sulfuric acid to produce aqueous calcium sulfate and water. • Pick out the chemical names in the reaction and turn them into chemical formulas calcium hydroxide + sulfuric acid → calcium sulfate + water • calcium hydroxide = Ca+2 OH-1, • Ca(OH)2 • Sulfuric acid = oxy acid, • Sulfuric was sulfate, H+1 SO4-2, • H2SO4 • Calcium Sulfate = Ca +2 SO4-2 • CaSO4 • Water = H2O • Skeleton Equation becomes: Ca(OH)2(s) + H2SO4(l) → CaSO4(aq) + H2O (l)

  30. Word Equations • Balancing Example 1 Ca(OH)2(s) + H2SO4(l) → CaSO4(aq) + H2O (l) Remember polyatomic rule only applies if it is the SAME on both sides. • Ca(OH)2(s) + H2SO4(l) → CaSO4(aq) + 2 H2O (l) • Ca(OH)2(s) + H2SO4(l) → CaSO4(aq) + H2O (l) • Ca 1 | Ca 1 • H 4 | H 2 4 • O 2 | O 1 2 • SO4 1 | SO4 1 • Ca 1 | Ca 1 • H 4 | H 2 • O 2 | O 1 • SO4 1 | SO4 1

  31. Word Equations • Example 2: Aluminum bromide and chlorine gas react to form aluminum chloride and bromine gas. • Don’t forget to balance charges and your 7 diatomic elements AlBr3 + Cl2 → AlCl3 + Br2 Now Balance Atoms like normal AlBr3 + Cl2 → AlCl3 + Br2 Al 1 | Al 1 Br 3 | Br 2 Cl 2 | Cl 3 2 AlBr3 + Cl2 → AlCl3 + 3 Br2 Al 1 2 | Al 1 Br 3 6 | Br 2 6 Cl 2 | Cl 3 2 AlBr3 + 3 Cl2 → 2 AlCl3 + 3 Br2 Al 1 2 | Al 1 2 Br 3 6 | Br 2 6 Cl2 6 | Cl3 6

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