Symbolic notation used by chemists to represent a chemical reaction. Chemical Equation Symbolic notation used by chemists to represent a chemical reaction.

In a chemical equation, everything to the left of the arrow Reactant – starting materials In a chemical equation, everything to the left of the arrow

In a chemical equation, everything to the right of the arrow Product – ending materials In a chemical equation, everything to the right of the arrow

Reactants  Products Arrow is read as “yields” or “produces” Equation format

Number in front of a formula. Tells how many formula units are present. Coefficient

Balanced chemical equation Demonstrates conservation of mass. Same # of each element on both sides. Balanced chemical equation

Chemical equations demonstrate conservation of mass, charge (& energy) Chemical equations demonstrate conservation of

1. Initial survey: count atoms of each element on each side. 2. Adjust numbers of atoms in each formula using coefficients. 3. Update atom counts after every change. How to balance equation?

Begin with the most complicated formula in the equation & work down to the simplest. Even numbers are easier to balance than odd numbers. If a polyatomic ion appears on both sides of the equation, treat it as a unit. Tricks for balancing

If you can’t balance an equation …. Check the formulas – odds are, at least one formula is incorrect! If you can’t balance an equation ….

Types of Reaction (5) Synthesis Decomposition Single Replacement Double Replacement Combustion Types of Reaction (5) Regents Level

Synthesis Reaction with 1 product Regents Level

Reaction with 1 reactant Decomposition Reaction with 1 reactant Regents Level

Single Replacement Reaction of 1 element with 1 compound to produce a new element & a new compound Regents Level

Reaction of 2 compounds to produce 2 new compounds. Double Replacement Reaction of 2 compounds to produce 2 new compounds. Regents Level

Reaction with oxygen. One of the reactants must be O2. Combustion Reaction with oxygen. One of the reactants must be O2. Regents Level

Decomposition AB  A + B A & B may be elements, compounds, or one of each Regents level

Synthesis A + B  AB A & B may be elements, compounds, or one of each Regents Level

Single Replacement, Cationic AX + B  A + BX A & B, metals or H Regents level

Single Replacement, Anionic X + AY  Y + AX X, Y = halogens Regents level

Double Replacement AX + BY  AY + BX Regents level

Combustion A + O2  Products Regents level

To predict if a single replacement reaction will occur, You must compare the reactivity of the free element with the corresponding element in the compound. Use Table J. If the stand-alone element is above the corresponding element in the compound, the reaction will take place. To predict if a single replacement reaction will occur,

2Na + MgCl2  2NaCl + Mg Na + MgCl2  ? Compare Na & Mg (both metals). Na is above Mg in Table J so the rxn occurs. 2Na + MgCl2  2NaCl + Mg Na + MgCl2  ?

F2 + MgCl2  MgF2 + Cl2 F2 + MgCl2  ? Compare F2 & Cl2 (both nonmetals). F2 is above Cl2 in Table J so the rxn occurs. F2 + MgCl2  MgF2 + Cl2 F2 + MgCl2  ?

Do the following reactions occur? Br2 + HF: Compare Br2 & F2 Mg + ZnCl2: Compare Mg & Zn Na + HCl: Compare Na & H2 Ag + LiBr: Compare Ag & Li No Yes Yes No

3 kinds of double replacement reactions 1) Those driven to completion by the formation of a precipitate Those driven to completion by the formation of a gas Those driven to completion by the formation of a molecular species, often H2O 3 kinds of double replacement reactions

3 kinds of double replacement reactions Two ionic compounds producing a solid precipitate. Two compounds producing a gas. Often two ionic compounds or an ionic compound with an acid or base. Two compounds producing a molecular species. For the 1st year: Acid + Base  water + … 3 kinds of double replacement reactions

To predict if a double replacement reaction will occur, You must determine the possible products & determine if the reaction goes to completion. The reaction goes to completion if a gas, a solid, or a small covalent molecule (H2O) is formed. Use Table F to determine if a precipitate (solid) is formed. To predict if a double replacement reaction will occur,

Write formulas for the possible products. Use Table F to determine if a precipitate (solid) is formed. Scan through table F looking for information about the solubility of the products. To predict if a double replacement reaction will produce a precipitate …

Ba(NO3)2(aq) + Fe2(SO4)3(aq)  ? Write formulas for the possible products: Fe(NO3)3 + BaSO4 Scan through table F looking for information about the solubility of the products: All nitrates are soluble so Fe(NO3)3 is aqueous. Most sulfates are soluble except when combined with Ag+, Ca2+, Sr2+, Ba2+, & Pb2+ so BaSO4 is solid. Ba(NO3)2(aq) + Fe2(SO4)3(aq)  ?

Stoichiometry Problems Use the relationship in the balanced equation to predict amounts (moles) consumed or produced. Use the equation coefficients to set up mole ratios. Stoichiometry Problems

General Procedure for Stoichiometry Problems 1st species into moles Mole conversion, species 1 to species 2, using coefficients from BCE 2nd species out of moles General Procedure for Stoichiometry Problems

Stoichiometry Problems May involve mole-mole relationships. For reactions involving all gases, coefficients in equation also give information about volume-volume ratios. Stoichiometry Problems

Factor-label method 4Na H2 = 2 moles of H2 2Na 2Na + 2H2O  2NaOH + H2 How many moles of hydrogen are produced when 4 moles of sodium reacts completely? Mole-mole problem

Proportion method The amount of Na is doubled, from 2 moles to 4 moles, so every other term in the equation is doubled. 2Na + 2H2O  2NaOH + H2 How many moles of hydrogen are produced when 4 moles of sodium reacts completely? Mole-mole problem

Factor-label method 3 L CH4 1 L CO2 = 3 L CO2 CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) How many liters of CO2 are produced when 3 liters of CH4 reacts completely? Volume-volume problem

Proportion Method The amount of CH4 is tripled, from 1 L to 3 L, so everything else is tripled. 3 liters of CO2 are produced! CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) How many liters of CO2 are produced when 3 liters of CH4 react completely?

Limiting Reactant Problems Give quantitative info about TWO reactants! Must find LR before doing stoichiometry! Limiting Reactant Problems

NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq) This is a limiting reactant problem because quantitative information about two reactants is given. A solution with 5.8 g of NaCl is mixed with a solution with 7.2 g of AgNO3. NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)

Limiting Reactant Problems Requirement: Must have a balanced chemical equation to begin a stoichiometry problem. Limiting Reactant Problems

Limiting Reactant Problems Two methods to determine LR 1) Tried and true 2) Grouping method Limiting Reactant Problems

90.0 g < 108. g so H2 = LR! Tried and True 10.0 g H2 1 mol H2 2 mol H2O 18 g H2O = 2 g H2 2 mol H2 1 mol H2O Calculation #1, based on H2 90. g Calculation #2, based on O2 108 g 96.0 g O2 1 mol O2 2 mol H2O 18 g H2O = 32 g O2 1 mol O2 1 mol H2O How many grams of water can be produced from 10.0 g H2 & 96.0 g O2? 2H2 + O2  2H2O

Grouping Method then stoichiometry. Find moles then groups. Smaller group is LR. # of groups = # of moles / coefficient 2.5 groups < 3 groups so H2 = LR! 2.5 groups 3 groups 2H2 + O2  2H2O 5 moles 3 moles 10.0 g 96.0 g How many grams of water can be produced from 10.0 g H2 & 96.0 g O2?

How many grams of water can be produced from 10.0 g H2 & 96.0 g O2? Make sure to use moles of LR, not groups, for given! For stoichiometry, use MOLES 5 mol H2 2 mol H2O 18 g H2O = 90 g H2O 2 mol H2 1 mol H2O H2 is LR – see previous slide! 2H2 + O2  2H2O How many grams of water can be produced from 10.0 g H2 & 96.0 g O2?

General Procedure for LR problems Into moles Identify LR Mole conversion, using coefficients from BCE. LR is “given” in mole conversion. Out of moles Excess reactant, two-step Calc amt of excess reactant consumed Subtract amt consumed from initial quantity General Procedure for LR problems

Excess reactant calculation, two-step! Use stoich conversion to find amt of O2 consumed in reaction Subtract amt consumed from initial amt Consider the previous problem: 5.00 mol 3.00 mol 2H2 + O2  2H2O 10.0 g 96.0 g Excess reactant calculation, two-step!

Excess reactant calculation, two-step! Given = LR Amount consumed 5 mol H2 1 mol O2 32 g O2 = 90 g O2 2 mol H2 1 mol O2 Subtract amt consumed from initial amt 96 g -90 g = 6 g O2 in excess. Consider the previous problem: 5.00 mol 3.00 mol 2H2 + O2  2H2O 10.0 g 96.0 g Excess reactant calculation, two-step!