Chemical Equations. shows the results of a chemical process
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1 Chemical Equations
2 Chemical Equations shows the results of a chemical process reactants (reagents) products coefficients the numbers in front of formulas in chemical equations gives the relative number of molecules taking part in a reaction
3 2H (g) 2 + O (g) 2 2H 2 O(g) Chemical bonds have been broken and new chemical bonds have been formed
4 Writing Chemical Equations 2H 2 (g) + O 2 (g) 2H 2 O (g) + 2 moles 1 mole 2 moles 4.04 g g g Parentheses show physical state of substances
5 Eample molar interpretation H 2 (g) + Cl 2 (g) 2HCl(g) Start: 1 mol 1 mol 0 Finish: mol
6 Eample mass interpretation 2H 2 + O 2 2H 2 O 4 g 32 g 36 g Start: 4g 32g 0 Finish: g
7 The law of conservation of mass requires that chemical equations must balance. What goes in Must come out
8 Quantitative Relationships
9 Stoichiometry the study of quantitative between amounts of reactants used and amounts of products formed how much reactant is needed to yield a certain amount of product?
10 Types of problems Mole - mole calculations Mass - mass calculations Limiting/Ecess reagent calculations
11 The mole method 1. Write and balance the equation. 2. Convert the given quantities into moles. 3. Use the coefficients in the balanced equation to relate the number of moles of known substances to the desired unknown one. 4. Convert to desired units. 5. Check your answer.
12 The mole method 1. Write and balance the equation. 2. Convert the given quantities into moles. 3. Use the coefficients in the balanced equation to relate the number of moles of known substances to the desired unknown one. 4. Convert to desired units. 5. Check your answer.
13 Stoichiometry Molar ratio of Y to X n y n Moles of X Moles of Y Mass of X X Y Mass of Y
14 Eample How many grams of nitrogen dioide can be formed by reaction of 1.44 g of nitrogen monoide with oygen? 2NO + O 2 2NO 2
15 Stoichiometry Molar ratio of Y to X n y n Moles of X Moles of Y 1.44 g of NO Mass of NO 2 2NO + O 2 2NO 2
16 Stoichiometry Molar ratio of Y to X n y n Moles NO Moles of Y 1.44 g of NO Mass of NO 2 2NO + O 2 2NO 2
17 Stoichiometry Molar ratio NO 2 to NO Moles NO Moles of Y 1.44 g of NO Mass of NO 2 2NO + O 2 2NO 2
18 Stoichiometry Molar ratio NO 2 to NO Moles NO Moles NO g of NO Mass of NO 2 2NO + O 2 2NO 2
19 Stoichiometry Molar ratio NO 2 to NO Moles NO Moles NO g of NO 2.21 g of NO 2 2NO + O 2 2NO 2
20 Eample 2NO + O 2 2NO g NO 1mol NO 30g NO 2mol NO 2 2mol NO 46g NO 2 1mol NO 2 = 2.21g NO 2
21 pg. 359 Problem 10a Calculate the number of moles of CS2 produced when 1.50 mol S 8 is used. 2 CH4 + S 8 2CS H 2 S 1.5 mol S 8 2 mol CS 2 = 3 mol CS 2 1 mol S 8
22 pg. 359 Problem 10b Calculate the number of moles of H2S produced when 1.50 mol S 8 is used. 2 CH4 + S 8 2CS H 2 S 1.5 mol S 8 4 mol H 2 S = 6 mol H 2 S 1 mol S 8
23 pg. 360 Problem 11 Titanium tetrachloride (TiCl4) is etracted from titanium oide using chlorine and carbon. TiO 2 + C + 2 Cl 2 TiCl 4 + CO 2 if you begin with 1.25 mol TiO2, what is the mass of Cl 2 needed? 1.25 mol TiO 2 2 mol Cl g Cl 2 1 mol TiO 2 1 mol Cl 2 = 177 g Cl 2
24 pg. 362 Problem 13 Air bags in cars can be inflated using the decomposition of sodium azide (NaN 3 ). 2 NaN 3 2 Na + 3 N 2 determine the mass of N2 produced if g of NaN 3 is decomposed? g NaN 3 1 mol NaN 3 3 mol N g NaN 3 2 mol NaN 3 28 g N 2 1 mol N 2 = g N 2
25 Limiting Reagents
26 Limiting Reagent Reactants are not always present (or available) in stoichiometric quantities. One reactant may be present in quantities such that it is completely consumed while ecess amounts of other reactants remain. - called limiting reactant or limiting reagent The limiting reagent will limit the amount of product produced.
27 Eample How many moles of MgCl 2 will be produced? Mg + Cl 2 MgCl 2 Start 1 mol 1 mol 0 Finish mol
28 Eample How many moles of MgCl 2 will be produced? Mg + Cl 2 MgCl 2 Start 1 mol 2 mol 0 Finish 0 1 mol 1 mol magnesium is the limiting reagent 1 mol of chlorine will be left unchanged
29 Limiting Reagent Molar ratio Y to X n y n Moles of X Moles of W Moles of Y Mass of X Mass of W W + X Y Mass of Y
30 Compare molar ratio W to X to their coefficients in balanced equation; identify LR Molar ratio Y to LR Moles of X Moles of W Moles of Y Mass of X Mass of W W + X Y Mass of Y
31 Eample Determine the limiting reagent and the amount of PI 3 produced when 6.00g P 4 reacts with 25.0g of I 2. P 4 + 6I 2 4PI 3 1mol P 6.00g P 4 4 =.0484mol P 4 124g P 4 1mol I 25.0g I 2 2 =.0984mol I 2 254g I 2
32 Eample cont... Determine how much I 2 would be needed to react completely with the available amount of P 4. if I have but I only have P 4 + 6I 2 4PI mol P 4 6mol I 2 1mol P 4 = I need 0.290mol I mol I 2 I 2 is the limiting reagent
33 Eample cont... the amount of PI 3 produced from the limiting reagent... P 4 + 6I 2 4PI mol I 2 4mol PI 3 6mol I 2 412g PI 3 1mol PI 3 = 27.0g PI 3
34 pg. 368 problem 20a the reaction between solid sodium and iron (III) oide is one in a series of reactions that inflates an automobile air bag Fe 2 O 3? 6Na + Fe 2 O 3 3Na 2 O + 2Fe if g Na and g Fe 2 O 3 are used in this reaction, determine: g Na a. the limiting reactant. 1 mol Na 23.0 g Na = mol Na 1 mol Fe 2 O g Fe 2 O g Fe 2 O 3 = mol Fe 2 O 3
35 pg. 368 problem 20a if g Na and g Fe 2 O 3 are used in this reaction, determine: a. the limiting reactant. 6Na + Fe 2 O 3 3Na 2 O + 2Fe if I have 1 mol Fe mol Na 2 O 3 = mol Fe2 O 3 6 mol Na I need mol Fe 2 O 3 is limiting but I only have
36 pg. 368 problem 20a if g Na and g Fe 2 O 3 are used in this reaction, determine: b. the reactant in ecess. 6Na + Fe 2 O 3 3Na 2 O + 2Fe mol Na in ecess 1 mol Fe 2 O 3 = mol Fe2 O 3 6 mol Na needed mol Fe 2 O 3 available
37 pg. 368 problem 20a if g Na and g Fe 2 O 3 are used in this reaction, determine: c. the mass of iron produced. 6Na + Fe 2 O 3 3Na 2 O + 2Fe 2 mol Fe g Fe mol Fe 2 O 3 1 mol Fe 2 O 3 1 mol Fe = g Fe produced
38 pg. 368 problem 20a if g Na and g Fe 2 O 3 are used in this reaction, determine: 6Na + Fe 2 O 3 6 mol Na mol Fe 2 O 3 1 mol Fe 2 O 3 = g Na used d. the mass of ecess reactant. 3Na 2 O + 2Fe 23.0 g Na 1 mol Na g Na g Na = 13.7 g Na left over
39 work sheet eample 3CaCO 3 + 2FePO 4 Ca 3 (PO 4 ) 2 + Fe 2 (CO 3 ) 3 if 100g calcium carbonate and 45 g iron(iii) phosphate are used in this reaction, determine: a. the limiting reactant. 100 g CaCO 3 1 mol CaCO g = 1 mol CaCO3 1 mol FePO 4 45 g FePO g = mol FePO 4
40 work sheet eample 3CaCO 3 + 2FePO 4 is limiting Ca 3 (PO 4 ) 2 + Fe 2 (CO 3 ) 3 if 100g calcium carbonate and 45 g iron(iii) phosphate are used in this reaction, determine: a. the limiting reactant. 1 mol CaCO 3 if I have mol FePO 4 but I only have 2 mol FePO 4 = 0.66 mol Fe2 O 3 3 mol CaCO 3 I need
41 work sheet eample 3CaCO 3 + 2FePO 4 Ca 3 (PO 4 ) 2 + Fe 2 (CO 3 ) 3 if 100g calcium carbonate and 45 g iron(iii) phosphate are used in this reaction, determine: b. grams of Ca 3 (PO 4 ) 2 formed? 1 mol Ca mol FePO 3 (PO 4 ) mol FePO g 1 mol Ca 3 (PO 4 ) 2 = 46.2 g Ca 3 (PO 4 ) 2 produced
42 work sheet eample 3CaCO 3 + 2FePO 4 Ca 3 (PO 4 ) 2 + Fe 2 (CO 3 ) 3 if 100g calcium carbonate and 45 g iron(iii) phosphate are used in this reaction, determine: b. grams of Fe 2 (CO 3 ) 3 formed? 1 mol Fe mol FePO 2 (CO 3 ) mol FePO g 1 mol Fe 2 (CO 3 ) 3 = 43.5 g Fe 2 (CO 3 ) 3 produced
43 work sheet eample 3CaCO 3 + 2FePO 4 Ca 3 (PO 4 ) 2 + Fe 2 (CO 3 ) 3 if 100g calcium carbonate and 45 g iron(iii) phosphate are used in this reaction, determine: c. grams of CaCO 3 unreacted? 3 mol CaCO mol FePO mol FePO g 1 mol CaCO 3 = 44.7 g CaCO 3 used 100 g g = 55.3 g CaCO 3 unused
44 Eample From the reaction between of 10.0g of Hg and 9.0g of Br 2. What mass of which reagent is left unreacted? Hg + Br g Hg 1 molhg 200.6gHg = HgBr 2 Hg is limiting molhg 9.0g Br 2 1 molbr 2 = molbr gBr 2
45 Hg + Br 2 HgBr molhg 1 molbr 2 = molbr 2 1 molhg moles of Br 2 needed to use up Hg available 159.8gBr molbr molbr 2 = 7.97 g Br 2 grams of Br 2 used 9.0g Br g Br 2 = 1.03 g Br 2 ecess
46 Reaction Yield
47 Theoretical yield the amount of product that would result if all the limiting reagent reacted Actual yield the amount of product actually obtained from the reaction Almost always less than the theoretical yield
48 Percent Yield Actual yield %yield = 100% Theoretical yield Determines how efficient a reaction is
49 Eample In a certain industrial operation g of TiCl 4 is reacted with g of Mg. (a) Calculate the theoretical yield of Ti in grams. (b) Calculate the percent yield if g are actually obtained. TiCl 4 + 2Mg Ti + 2MgCl 2
50 Calculate theoretical yield g TiCl g Mg 1mol TiCl 4 = mol 187.7g TiCl 4 TiCl 4 I have 1mol Mg = mol 24.31g Mg Mg if I have 2mol Mg mol TiCl 4 = mol Mg 1mol TiCl 4 I need there is more than enough Mg TiCl 4 is limiting
51 g TiCl 4 1mol TiCl 4 1mol Ti 187.7g TiCl 4 1mol TiCl g Ti 1mol Ti = g Ti 100% = g Ti g Ti Actual yield %yield = 100% = 88.6% theoretical Theoretical yield
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54 another method Determine the limiting reagent and the amount of PI 3 produced when 6.00g P 4 reacts with 25.0g of I 2. P 4 + 6I 2 4PI g P 4 1mol P 4 124g P 4 =.0484mol P g I 2 1mol I 2 254g I 2 =.0984mol I 2
55 another method P 4 + 6I 2 4PI mol I 2 = mol P 4 The actual I 2 /P 4 ratio is less than the stiochiometic ratio 6 mol I 2 = 6 So there is not enough I2 1 mol P to 4 I 2 is the limiting reagent react with all the P 4
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