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Lets take a moment… sit back… relax… and review some previously learned concepts… Review of Classifying and Balancing Chemical Equations.

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Presentation on theme: "Lets take a moment… sit back… relax… and review some previously learned concepts… Review of Classifying and Balancing Chemical Equations."— Presentation transcript:

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4 Lets take a moment… sit back… relax… and review some previously learned concepts… Review of Classifying and Balancing Chemical Equations

5 5 main types of chemical reactions: 1. Formation/Synthesis 2. Decomposition/De-formation 3. Single Replacement 4. Double Replacement 5. Hydrocarbon Combustion 6. Other Combustions

6 1. Formation/Synthesis Reactions Reaction in which 2 or more pure elements combine to form a new compound. General Equation: A + B  AB Element + Element  Compound Example: Na + Cl  NaCl (Sodium) (Chlorine) (Sodium Chloride) SALT

7 2. Decomposition/De-Formation Reaction in which a compound breaks down into the base elements that formed it. General Equation: AB  A + B Compound  Element + Element Example: H 2 O (l)  H 2(g) + O 2(g) (Water) (Hydrogen)(Oxygen)

8 3. Single Replacement Reaction in which an element and a compound react. The “partners” are traded and the elements that make up the compound and lone element are substituted. General Equation: ABCACB AB +C  AC + B Compound Element CompElemen Compound + Element  Compound + Element Example: NaCl (s) LiCl (s) Na (s) NaCl (s) + Li (s)  LiCl (s) + Na (s) (Salt) (Lithium) (Lithium Chloride) (Soduim)

9 4. Double Replacement Reaction in which two compounds react and “switch partners”. The elements that make up each compound are swapped to form two new compounds. General Equation: ABCDADCB AB +CD  AD + CB Compound CompoundCompCompound Compound + Compound  Compound + Compound Example: NaCl (s) LiCl (s) NaBr (s) NaCl (s) + LiBr (s)  LiCl (s) + NaBr (s) (Salt) (Lithium Bromide) (Lithium Chloride) (Sodium Bromide)

10 5. Hydrocarbon Combustion Reaction in which a compound containing Hydrogen and Carbon are burned to produce H 2 O (g) and CO 2(g). General Equation: C X H X O 2(g) H 2 O (g) CO 2(g) C X H X +O 2(g)  H 2 O (g) + CO 2(g) C x H x OxygenWaterCarbon Dioxide C x H x + Oxygen  Water + Carbon Dioxide Example: CH 4(g) H 2 O (g) CO 2(g) CH 4(g) + O 2(g)  H 2 O (g) + CO 2(g) (Methane) (Oxygen) (Water Vapour) (Carbon Dioxide)

11 6. Other Combustions Reaction in which a compound OR element is burned forming its most common oxide. General Equation: XO 2(g) XO (g) X +O 2(g)  XO (g) X OxygenXO X + Oxygen  XO Example: XXO X+ O 2(g)  XO

12 Steps to Balance a Chemical Equation 1)Write out the correct chemical formulae for the products and the reactants. Be sure to include all states of matter! 2)Balance the atoms or ion present in the greatest number. You may do this by finding the lowest common multiple of the two. 3)Continue to systematically balance the rest of the atoms or ions. 4)Check the final equation. Make sure every type of atom or ion balances.

13 The coefficients that are placed in front of the compounds and elements represent the mole-to-mole ratio of the reactants and products. Ex. 2NaCl (aq) + MgSO 4(aq) → Na 2 SO 4(aq) + MgCl 2(s) – 2 moles of sodium chloride reacts with one mole of magnesium sulfate to produce one mole of sodium sulfate and one mole of magnesium chloride When you are making quantitative calculations, please use the mole-to-mole ratio in your calculations :

14 RXN Limitations RXNS don’t communicate temp. and pressure RXNS don’t communicate progress or process RXNS don’t communicate measurable quantities of substances (can’t really weigh or calculate moles, atoms, molecules, ions, etc. without a calculation first)

15 RXN Assumptions RXNS are spontaneous RXNS are fast RXNS are quantitative (more than 99% complete) RXNS are stoichiometric (simple, whole number ratio of chemical amounts of reactants and products

16 Net Ionic Equations Net ionic equations are useful in that they show only those chemical species participating in a chemical reaction The key to being able to write net ionic equations is the ability to recognize monoatomic and polyatomic ions, and the solubility rules.

17 Net Ionic Equations Pb(NO 3 ) 2(aq) + 2HCl (aq)  PbCl 2(s) + 2 HNO 3(aq) Step 1: Step 2: Dissociate (split up) and soluble ionic compounds AND ionize (split up) all Strong Acids. Pb 2+ (aq) 2NO 3 - (aq) ++ 2 H 3 O + (aq) + 2 Cl - (aq)  PbCl 2(s) + 2 H 3 O + (aq) +2NO 3 - (aq)

18 Net Ionic Equations Pb 2+ (aq) 2NO 3 - (aq) H 3 O + (aq) + 2 Cl - (aq)  PbCl 2(s) + 2 H 3 O + (aq) + 2NO 3 - (aq) Total Ionic Equation Step 3: Cancel out the spectator ions. (Ions that are the same on both sides) Step 4: Write down whatever is left over after canceling. Simplify balancing if necessary Pb 2+ (aq) + 2 Cl - (aq)  PbCl 2(s) Net Ionic Equation

19 BaCl 2+Na 2 SO 4 (aq) (aq)  What type of reaction? Double Replacement NaCl + BaSO 4 ****Have to use solubility table to determine the state (aq) or (s)**** (aq)(s) 2 Complete Balanced Chemical Equation Step 1: DONE

20 BaCl 2+Na 2 SO 4 (aq) (aq)  NaCl + BaSO 4 (aq)(s) 2 Complete Balanced Chemical Equation Step 2: Dissociate soluble ionic and ionize strong acids. Ba 2+ (aq) 2Cl - (aq) 2Na + (aq) SO 4 2- (aq) 2Na + (aq) 2Cl - (aq) BaSO 4(s)

21 Step 2: DONE Ba 2+ (aq) 2Cl - (aq) 2Na + (aq) SO 4 + (aq) 2Na + (aq) 2Cl - (aq) BaSO 4(s) ++ +  + + Step 3: Cancel out the spectator ions. (Ions that are the same on both sides)DONE Step 4: Write down whatever is left over after canceling. Simplify balancing if necessary. Ba 2+ (aq) + SO 4 + (aq)  BaSO 4(s) DONE

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23 Gravimetric Stoichiometry: using the balancing numbers and (Need/Got) The process of calculating the masses involved in a chemical reaction using the balancing numbers and (Need/Got). Starting Steps for Starting stoichiometry: 1.Balanced Chemical Equation. 2.Write information…mols….masse under the chemical/element that it relates to. 3.***Convert information given in question into mols 3.***Convert information given in question into mols.*** 4.Find the chemical species the question wants to know about and put a N for (need to find) above it. 5.Put a G for (got information about this) over the species you have information about from the question.

24 Step 1: Write a Balanced Chemical Equation for the reaction. Fe 2 O 3(s) + CO (g)  Fe (s) + CO 2(g) m = ? m = 100.0g Step 2: Write the information from the question under the correct element/compound Step 3: Convert information to mols. n = ? n = m M n = g g/mol n =1.79 mol

25 Fe 2 O 3(s) + CO (g)  Fe (s) + CO 2(g) m = ? m = 100.0g Step 4: Put an N over the species you NEED to find out about. Step 5: Put a G over the compound you know information about. n = m M n = g g/mol n = 1.79 mol N G

26 Hydrogen gas and Oxygen gas react to produce water. How much hydrogen gas would be needed to produce g of water? Step 1: Write a Balanced Chemical Equation for the reaction. Step 2: Write the information from the question under the correct element/compound Step 3: Convert information to mols. Step 4: Put an N over the species you NEED to find out about. Step 5: Put a G over the compound you know information about. 2 H 2(g) + 1 O 2(g)  2 H 2 0 (l) m = gm = ? n = mol N G

27 Gravimetric Stoichiometry: using the balancing numbers and (Need/Got) The process of calculating the masses involved in a chemical reaction using the balancing numbers and (Need/Got). FINISHING Steps for FINISHING stoichiometry: 6. Convert mols of got (G) to mols of need (N) by multiple by the conversion factor (N/G). 7. Convert the mols of need (N) into a mass by multiplying by the molar mass (M) of the needed compound.

28 Fe 2 O 3(s) + CO (g)  Fe (s) + CO 2(g) m = ? m = 100.0g n =1.79 mol N G Step 6: Convert mols of got (G) to mols of need (N) by multiple by the conversion factor (N/G). ( ) NGNG 1212 n FE x (1/2) = n Fe2O3 (1.79mol) X (1/2) = n Fe2O mol = n Fe2O3 n =0.895 mol

29 Fe 2 O 3(s) + CO (g)  Fe (s) + CO 2(g) m = ? m = 100.0g n = mol N G Step 7: Convert mols of need (N) into mass of Need (N) by multiplying by N’s Molar Mass. ( ) 1212 M Fe2O3(s) = n =0.895 mol n = m M m = nM m = (0.895 mol)( ) 2 x g/mol 3 x g/mol g/mol m = g

30 Hydrogen gas and Oxygen gas react to produce water. How much hydrogen gas would be needed to produce g of water? 2 H 2(g) + 1 O 2(g)  2 H 2 0 (g) m = gm = ? n = mol N G Step 6: Convert mols of got (G) to mols of need (N) by multiple by the conversion factor (N/G). Step 7: Convert the mols of need (N) into a mass by multiplying by the molar mass (M) of the needed compound. m = g

31 Excess and Limiting Reagents Excess and limiting reagents refer to the reactant that will run out first and stop more product from forming. + = 24 Crackers7 Pieces of Cheese Which will run out first? Limiting Reagent Excess Reagent

32 Excess and Limiting Reagents 12.4 g of Hydrogen gas reacts with 12.4 g of Oxygen gas to form liquid water. Determine the excess and limiting reagent. Calculate the maximum amount of water that could be produced by reacting these two gases together. H 2(g) + O 2(g)  H 2 O (g) m = 12.4 g m = ? n = 6.138….mol n = m M n = m M n = mol N G ***From here on it’s like a double stoichiometry to see which can make more H 2 O (l )***

33 Excess and Limiting Reagents 12.4 g of Hydrogen gas reacts with 12.4 g of Oxygen gas to form liquid water. Determine the excess and limiting reagent. Calculate the maximum amount of water that could be produced by reacting these two gases together. H 2(g) + O 2(g)  H 2 O (g) m = 12.4 g m = ? n = 6.138….moln = mol N G ( ) NGNG 2222 H2 n = 6.138…

34 Excess and Limiting Reagents 12.4 g of Hydrogen gas reacts with 12.4 g of Oxygen gas to form liquid water. Determine the excess and limiting reagent. Calculate the maximum amount of water that could be produced by reacting these two gases together. H 2(g) + O 2(g)  H 2 O (g) m = 12.4 g m = ? n = 6.138….moln = mol N G ( ) NGNG 2121 H2 n = 6.138… O2 n = mol Limiting Excess

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36 Percent Yield In every reaction you can use stoichiometry to calculate the theoretical amount of product that could be made. (Maximum or Total). When you actually do an experiment, the actual amount that you are able to make is called the actual amount. % Yield = Actual Amount Theoretical Amount X 100%

37 mols Stoichiometry ALWAYS involves mols and a balanced chemical equation. H 2(g) + O 2(g)  H 2 O (g) What information can get us n? What formula’s do we know that have n in them?

38 H 2(g) + O 2(g)  H 2 O (g) n = m M Use this when we turn a mass into mols and do stoichiometry. ALREADY DID THIS. Called Gravimetric Stoichiometry. STP SATP 22.4 mol/L24.8 mol/L n = 1 mol V = 22.4 L n = 1 mol V = 24.8 L ***Because we can calculate mols we can then do stoichiometry just like before.***

39 C 3 H 8(g) + O 2(g)  CO 2(g) + H 2 O (g) Even Number 51

40 C 3 H 8(g) + O 2(g)  CO 2(g) + H 2 O (g) m = 275g V = ? N G n = m M n = n = 275g g/mol 6.23 mol ( ) NGNG 5151 n = 31.2 mol 22.4 L 1 mol n STP = = X 31.2 mol X = 698 L V = 698 L

41 (YOU TRY) Na (s) + H 2 O(l)  H 2(g) + NaOH (s) N G m = ? V = 20.0 L 24.8 L 1 mol n SATP = = 20.0 L X X = mol n = 0.806…mol ( ) 2121 n = 1.61…mol n = m M m= nM m= (1.61mol)(22.99g/mol) m= 37.1 g

42 NH 3(g) N 2(g)  + H 2(g) V = ? 21 3 N P = 450 kPa T = 80°C T = K

43 NH 3(g) N 2(g)  + H 2(g) V = ? 21 3 N P = 450 kPa T = K m = 7.5 Kg m = 7500 g G

44 NH 3(g) N 2(g)  + H 2(g) V = ? 21 3 N P = 450 kPa T = K m = 7500 g G R = Have to calculate mols to do stoich. n = m M n = 7500g 2.02 g/mol n = 3700 mol n =( ) NGNG mol

45 V = ? P = 450 kPa T = K R = n = mol PV = nRT V = ( mol)(8.314)(333.15) (450 kPa) V = L of NH 3 V = 16 KL of NH 3

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47 mols Stoichiometry ALWAYS involves mols and a balanced chemical equation. Solution Stoichiometry: Using the principles that govern concentrations to solve for n (mols). Once you have moles the stoichiometry is the same. C = n V ***Once you get mols…the rest is the same.***

48 H 2 SO 4(aq) + KOH (aq)  H 2 O (aq) + K 2 SO 4(aq)

49 H 2 SO 4(aq) + KOH (aq)  H 2 O (aq) + K 2 SO 4(aq)

50 H 2 SO 4(aq) + KOH (aq)  H 2 O (aq) + K 2 SO 4(aq) V = L V = L C = mol/L C = n V n = CV N G n = (0.150 mol/L)( L) n = mol ( ) NGNG 1212 n = C = ?

51 H 2 SO 4(aq) + KOH (aq)  H 2 O (aq) + K 2 SO 4(aq) V = L V = L C = mol/L C = n V N G n = C = ? C = mol 0.010L C = mol/L

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