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Symbolic notation used by chemists to represent a chemical reaction.

Symbolic notation used by chemists to represent a chemical reaction. Chemical Equation. In a chemical equation, everything to the left of the arrow. Reactant – starting materials. In a chemical equation, everything to the right of the arrow. Product – ending materials. Equation format.

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Symbolic notation used by chemists to represent a chemical reaction.

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

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

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

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

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

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

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

  8. 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?

  9. Tricks for balancing 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.

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

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

  12. Reaction with 1 product Synthesis Regents Level

  13. Reaction with 1 reactant Decomposition Regents Level

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

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

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

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

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

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

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

  21. AX + BY  AY + BX Double Replacement Regents level

  22. A + O2 Products Combustion Regents level

  23. 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.

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

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

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

  27. 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

  28. 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 + …

  29. 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.

  30. To predict if a double replacement reaction will produce a precipitate … 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.

  31. 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.

  32. 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.

  33. 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

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

  35. 4Na H2 = 2 moles of H2 2Na + 2H2O  2NaOH + H2How many moles of hydrogen are produced when 4 moles of sodium reacts completely? 2Na Factor-label method Mole-mole problem

  36. 2Na + 2H2O  2NaOH + H2How many moles of hydrogen are produced when 4 moles of sodium reacts completely? Proportion method The amount of Na is doubled, from 2 moles to 4 moles, so every other term in the equation is doubled. Mole-mole problem

  37. 1 L CH4 CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)How many liters of CO2 are produced when 3 liters of CH4 reacts completely? Factor-label method 3 L CH4 1 L CO2 = 3 L CO2 Volume-volume problem

  38. CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)How many liters of CO2 are produced when 3 liters of CH4 react completely? 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!

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

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

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

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

  43. 90.0 g < 108. g so H2 = LR! Calculation #1, based on H2 90. g 2H2 + O2 2H2O 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 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 How many grams of water can be produced from 10.0 g H2 & 96.0 g O2?

  44. 2H2 + O2 2H2O 2.5 groups < 3 groups so H2 = LR! 2.5 groups 3 groups 5 moles 3 moles Grouping Method then stoichiometry. Find moles then groups. Smaller group is LR. # of groups = # of moles / coefficient 10.0 g 96.0 g How many grams of water can be produced from 10.0 g H2 & 96.0 g O2?

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

  46. 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

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

  48. 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!

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