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Unit 5: Chemical Reactions
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Law of Conservation of Mass
Atoms are neither created nor destroyed during a chemical reaction Elements are present in the same amount before and after a reaction they simply move around, breaking old bonds and creating new ones
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Chemical Reactions Chemical reactions occur when two or more compounds interact, breaking bonds and creating new ones yielding products that are different from the substances being reacted
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Chemical Reactions Many ways to know reaction has occurred
Temperature change Colour change Bubbles Odour produced Precipitate or new solid produced
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Chemical Equations Used to summarize a chemical reaction
“+” means “reacts with” “” means “produces” Substances to the left of the arrow are reactants Substances to the right of the arrow are products
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Chemical Equations Coefficients are the big numbers used to tell the ratio of mols of each reactant and product Subscripts are small numbers used to tell the ratio of mols of each element
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Chemical Equations
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Chemical Equations Subscripts are also used to show the state of reactants and products (s) = solid (l ) = liquid (g) = gas (aq) = aqueous (dissolved in water), used for solutions, acids, and bases
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Balancing Chemical Equations
Remembering the Law of Conservation of Mass we have to have the same amount of each element is our products as we have in our reactants We using “balancing” to make sure we follow this law
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Balancing Chemical Equations
In reactions all subscripts make each compound what it is We do not change small numbers in an equation We change coefficients (big numbers) in order to balance equations We have to balance equations so we have the lowest whole number ratio between all reactants and products
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Balancing Chemical Equations
Balance the following equations: H2(g) + O2(g) H2O(l ) KClO3(aq) KCl(s) + O2(g) C2H6(g) + O2(g) CO2(g) + H2O(g)
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Writing Chemical Reactions
In a precipitation reaction, sodium hydroxide solution is mixed with iron (II) chloride solution. Sodium chloride solution and insoluble iron(II) hydroxide are produced. Write a balanced chemical equation including the state symbols.
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Writing Chemical Reactions
Calcium hydroxide, a base reacts with phosphoric acid to yield solid calcium phosphate and liquid water. Determine the balanced equation.
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Types of Reactions Combination Decomposition
Single displacement/replacement Double displacement/replacement Neutralization Precipitate formation Combustion
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Combination Reactions
When two or more substances react to form one product A + B AB For this course Two elements combining Formation of hydrates Non-metal and metal oxides and water
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Two Element Combination Reactions
Two elements combine to create a new compound Predict the products of the following: C(s) + O2(g) 2Mg(s) + O2(g) N2(g) + 3H2(g)
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Hydrate Combination Reactions
Hydrates as stated before are chemical compounds that contain water or its parts For this course we will mainly deal with hydrates of inorganic compounds (ionic compounds) We can add water to compounds in a lattice work structure surrounding the central compound
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Hydrate Combination Reactions
CoCl2 + 6H2O CoCl2 •6H2O
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Non-metal Oxide and Water Combination Reactions
Always forms an acid when oxide is soluble in water Acids begin with H Predict the products of the following: SO2 + H2O Cl2O5 + H2O
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Metal Oxide and Water Combination Reactions
Always forms a base when oxide is soluble in water Contain hydroxide ion (-OH) Predict the products of the following: Na2O + H2O MgO + H2O
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Decomposition Reactions
One substance undergoes a reaction to produce two or more other substances AB A + B For this course: Compound breaks down into its elements Hydrates Carbonates
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Decomposition of Compound into its Elements
One compound (usually by heat) is forced to break down into stable states of the elements that make it up Predict the products of the following: HgO MgCl2
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Decomposition of Hydrates
We can heat hydrated substances to remove water molecules The result is an “anhydrous” form of the compound This can sometimes change the colour of the substance
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Decomposition of Hydrates
CuSO4 •5H2O CuSO4 + 5H2O
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Decomposition of Carbonates
All carbonates break down to yield an oxide and CO2 Predict the products of the following: CaCO3 Na2CO3
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Single Displacement Reactions
AB + C AC + B We must use the activity series to determine whether a reaction will take place or not The strongest/most reactive cation will create a bond If the most reactive cation is already bonded then no reaction will take place
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Single Displacement Reactions
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Single Displacement Reactions
Those atoms closer to the top of the activity series (more reactive will replace lower elements in bonds.
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Single Displacement Reactions
Determine whether the following reactions occur and write the products if a reaction is present: AlCl3 + Ba ZnI2 + Pb
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Neutralization Reactions
Reactions involving acids and bases as the reactants These reaction will yield water and a salt as their products “salt” is any ionic compound whose cation comes from a base and anion comes from an acid
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Neutralization Reactions
HCl(aq) + NaOH(aq) H2O(l ) + NaCl(aq)
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Double Displacement Reactions
AB + CD AD + BC Use solubility rules to determine if a precipitate is formed No precipitate = no reaction Precipitate is a solid that is form by two or more chemicals reacting together, sometimes changes the colour of the solution
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Double Displacement Reactions
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Double Displacement Reactions
Determine for the following if a reaction occurs, if so state the products and their states. AgNO3(aq) + K3PO4(aq) CuF(aq) + AlCl3(aq) AlCl3(aq) + NaI(aq)
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Carbonates and Bicarbonates with Acids
React to form carbonic acid (H2CO3) Carbonic acid is unstable and will continue to react This gives H2O and CO2 as final products HCl(aq) + NaHCO3(aq) NaCl(aq) + H2CO3(aq) H2CO3(aq) H2O(l) + CO2(g)
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Combustion Reactions Combustion is burning (we always add O2(g) to whatever is being combusted) In this course we assume there is always complete combustion Combustion of hydrocarbons (compounds with only H and C) always produces CO2(g) and H2O(g) All you need to do is balance (no guesswork with products)
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Combustion Reactions Predict the products of the following and balance the final equation C4H10 + O2(g) C7H16 + O2(g)
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Net Ionic Equations For reactions in aqueous solution
Useful to show whether dissolved substances are present mostly as ions or molecules Break down each reactant and product to the ions that make them up (if they are truly molecular and do not break up into ions write them as the whole compound)
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Net Ionic Equations Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)
Becomes: Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) PbI2(s) + 2K+(aq) + 2NO3-(aq) The above is the complete ionic equation All soluble strong electrolytes are shown as ions Remember if they aren’t soluble they are left PbI2(s)
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Net Ionic Equations Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) PbI2(s) + 2K+(aq) + 2NO3-(aq) K+(aq) and NO3-(aq) appear on both sides of the equation in the same form These are called spectator ions Don’t participate in the reaction When spectator ions are omitted we have the net ionic equation Pb2+(aq) + 2I-(aq) PbI2(s)
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Net Ionic Equations Write Net Ionic Equations for the following
AgNO3(aq) + KCl(aq) AgCl(s) + KNO3(aq) Mg(NO3)2(aq) + Na2CO3(aq) MgCO3(s) + NaNO3(aq)
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Oxidation Reduction Reactions
Redox reactions Reactions where electrons are transferred from one reactant to another
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Redox Reactions LEO the lion says GER Loss of Electrons is Oxidation
Gain of Electrons is Reduction When one substance is oxidized another substance must be reduced Electrons have to go somewhere
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Redox Reactions Each atom in a neutral substance of ion is assigned an oxidation number (oxidation state) Oxidation states allow us to keep track of electrons being gained or lost Oxidation numbers range from -4 to +7
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Oxidation Numbers Oxidation numbers of atoms in covalently bonded compound depend on electronegativity More electronegative atoms carrying a more negative oxidation state Written with +/- before the number (S-2) Charge is written +/- after the number (S2-)
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Oxidation Numbers 1. Atoms in their natural state have oxidation number of 0 Each H in H2 has oxidation state of 0 Each S in S8 has oxidation state of 0 Each P in P4 has oxidation state of 0 Solid metals (ie. Fe(s))have oxidation state of 0
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Oxidation Numbers 2. Monoatomic ions’ oxidation number = ionic charge K+ = oxidation state +1 S2- = oxidation state -2
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Oxidation Numbers 3. Non-metals usually have negative oxidation state
Some exceptions O usually -2 Exception = peroxides (contain O22- ion, given -1) H usually +1 with non-metals, -1 with metals F is -1 in all compounds Other halogens = -1 in most binary compounds When combined with O have positive oxidation states
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Oxidation Numbers Sum of oxidation numbers in neutral compound = 0
Sum of oxidation numbers in polyatomic ion =charge of the ion H3O+ = 3(+1) + 1(-2) = +1 = charge of ion
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Oxidation Numbers What is the oxidation number of sulfur in the following: H2S S8 SCl2 Na2SO3 SO42-
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Redox Reaction In order for a reaction to be considered redox there needs to be a transfer of electrons This means that there must be a change in oxidation state Remember, when an electron is lost by one substance another substance must gain that electron We cannot have oxidation without reduction
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Finding Redox Reactions
Determine which of the following are redox: Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) Na2CO3(s) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)
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Redox Agents Oxidizing agent undergoes reduction
Causes another species to be oxidized Reducing agent undergoes oxidation Causes another species to be reduced
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Redox Agents Determine the oxidizing agent and reducing agent in each example: Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)
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Stoichiometry We can use balanced chemical equations to convert from one substance to another This can help us determine how much of our product we can expect to get Remember, equations must be balanced!!!!
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Stoichiometry Coefficients in an equation tell us the mol ratio
Compares mols of one substance to another To convert between two different substances we must be in mols
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Stoichiometry Given the following equation how many mols of ZnBr2 will be produced from mols of solid Zn? Zn(s) + 2HBr(aq) ZnBr2(aq) + H2(g)
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Stoichiometry Given the following equation how many mols of solid Pb will be produced from 33.3gPb(NO3)2? Mn(s) + Pb(NO3)2(aq) Mn(NO3)2(aq) + Pb(s)
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Stoichiometry Given the following equation how many grams of H2 can be made with 6.5gHCl? Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)
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Limiting Reagents (Reactants)
Imagine you are making sandwiches using bread and cheese For every 2 pieces of bread you need one piece of cheese Bd = bread Ch = cheese Bd2Ch = sandwich
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Limiting Reagents (Reactants)
The following equation represents your sandwich making: 2Bd + Ch Bd2Ch If you have 10 slices of bread and 7 slices of cheese you would only be able to make 5 sandwiches Once you run out of bread you can’t make any more
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Limiting Reagents (Reactants)
In the lab once we run out of one of our reactants we can’t make any more products The first reactant to run out is known as the limiting reagent or limiting reactant In calculations we use the limiting reagent for calculating the amount of product that will be made Remember to balance your equation!
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Limiting Reagents (Reactants)
Given the following reaction how many mols NH3 can be formed from 3.0mol N2 and 6.0 mol H2? N2(g) + 3H2(g) 2NH3(g)
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Limiting Reagents (Reactants)
How many grams of H2O can be formed according to the following equation if 150g H2 are reacted with 1500g O2? 2H2(g) + O2(g) 2H2O(g)
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Percent Yield Comparison between theoretical yield and actual yield
Theoretical yield = calculated amount of product From amount of reactants started with (remember use limiting reagent) Actual yield = from doing experiment Must be given in problem or found in lab
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Percent Yield %𝑦𝑖𝑒𝑙𝑑= 𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 x 100%
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Percent Yield Student preformed the following experiment in lab. Using 3.00gTi and 6.00gCl2 students were able to produce 7.7gTiCl4. Determine the percent yield for this experiment. Ti(s) + Cl2(g) TiCl4(l)
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