Unit 6: Balancing Chemical Equations, Chemical Reactions & Stoichiometry The GowerHour

VII. Stoichiometric “road map” (Use the balanced chemical equation) aA + bB cC + dD Mol to Mol Ratio Using the Coefficients from the balanced chemical equation Mass A Mass B Molar mass Molar mass Mol A Mol B 6.022 x 1023 6.022 x 1023 Atoms Molecules Formula Units A Atoms Molecules Formula Units B

atoms; particles; molecules; formula units 10.1 D. Converting Number of Particles to Moles 1. One mole (mol) of a substance is 6.022  1023 representative particles of that substance and is the SI unit for measuring the amount of a substance. 2. The number of representative particles in a mole, 6.022  1023, is called Avogadro’s number. atoms; particles; molecules; formula units

3. Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, tools, and garden furniture. How many moles of magnesium is 1.25 x 1023 atoms of magnesium? 1 mol Mg atoms = 6.022 x 1023 atoms Mg 1 mol Mg atoms 1.25 x 1023 atoms Mg X 6.022 x 1023 atoms Mg = 0.208 mol Mg = 2.08 x 10-1 mol Mg

atoms; particles; molecules; formula units 10.1 E. Converting Moles to Number of Particles 1. 1.56 moles of methanol, CH3OH, contain how many molecules? atoms; particles; molecules; formula units 1 mol molec CH3OH = 6.022 x 1023 molec CH3OH 6.022 x 1023 molec CH3OH 1.56 mol CH3OH x 1 mol CH3OH = 9.40 x 1023 molec. CH3OH

Calculating Molecular Mass Calculate the molecular mass of magnesium carbonate, MgCO3. 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g/mol

2. Aluminum satellite dishes are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass of 9.45 mol of aluminum oxide? 1 mol Al2O3 = 101. 96 g Al2O3 101. 96 g Al2O3 9.45 mol Al2O3 x = 964 g Al2O3 1 mol Al2O3 These aluminum satellite dishes at the National Radio Astronomy Observatory near Soccoro, New Mexico are naturally protected from corrosion by the formation of a thin film of aluminum oxide (Al2O3).

4. When iron is exposed to air, it corrodes to form red-brown rust 4. When iron is exposed to air, it corrodes to form red-brown rust. Rust is iron(III) oxide (Fe2O3). How many moles of iron(III) oxide are contained in 92.2 g of pure Fe2O3? 1 mol Fe2O3 = 159.70 g Fe2O3 1 mol Fe2O3 92.2 g Fe2O3 x = 0.577 mol Fe2O3 159.70 g Fe2O3 = 5.77 x 10-1 mol Fe2O3 Rust weakens an iron chain.

Everything must go through Moles!!! Calculations molar mass Avogadro’s number Grams Moles particles Everything must go through Moles!!!

mass IV. Stoichiometry = The _____ relationships among reactants and products in chemical reactions. A chemical equation is a statement of ___________ fact. On the left side of the reaction are the _________, and on the right side the _________ of the reaction. Since no atoms are created nor destroyed in a ___________ chemical reaction, the equation must be _________. This means that the combined weight of the reactants is exactly ______ to the combined weight of the products. In terms of chemical laws, this is the Law of _____________ of matter (__________). experimental reactants products non-nuclear balanced equal conservation Lavoisier

V. Limiting and Excess Reagents A. In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. B. The _____________ is the reagent that determines the amount of product that can be formed by a reaction. C. In the reaction of nitrogen and hydrogen, hydrogen is the limiting reagent. Nitrogen is the reagent that is not completely used up in the reaction. The reagent that is not used up is called the ____________. limiting reagent excess reagent

D. Analogy: How to make a cheese sandwich.   2 slices of bread + 1 slice of cheese → 1 cheese sandwich   If you have 8 slices of bread and 6 slices of cheese, how many sandwiches can you make? __ (theoretical yield)   What is the limiting reagent? ______   What is the excess reagent? _______   How much of the excess reagent is left at the end of the rxn? ______________ 4 bread cheese 2 slices cheese

VI. Percent Yield A. The percent yield is a measure of the __________ of a reaction carried out in the laboratory. B. A batting average is actually a percent yield. efficiency

theoretical C. The ____________ yield is the maximum amount of product that could be formed from given amounts of reactants. D. In contrast, the amount of product that actually forms when the reaction is carried out in the __________ is called the actual yield. E. The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent. laboratory

VII. Stoichiometric “road map” (Use the balanced chemical equation) aA + bB cC + dD Mol to Mol Ratio Using the Coefficients from the balanced chemical equation Mass A Mass B Molar mass Molar mass Mol A Mol B 6.022 x 1023 6.022 x 1023 Atoms Molecules Formula Units A Atoms Molecules Formula Units B

Examples: A. Calcium reacts with oxygen to form calcium oxide. If you are given 80.0 g of calcium, (1) Calculate the theoretical yield of calcium oxide. (2) How many moles of your excess reagent (ER) are used? LR Ca + O2 CaO 2 2 ER 1 mol Ca 2 mol CaO 56.08 g CaO (1) 80.0 g Ca x x x 40.08 g Ca 2 mol Ca 1 mol CaO LR = 112 g CaO Theoretical Yield ER 1 mol Ca 1 mol O2 = 0.998 mol O2 (2) 80.0 g Ca x x 40.08 g Ca 2 mol Ca LR ER used

B. Nitrogen and oxygen react to form dinitrogen pentoxide B. Nitrogen and oxygen react to form dinitrogen pentoxide. If 112 g of nitrogen react with unlimited oxygen, (1) how many moles of the excess reagent (ER) react and (2) how many grams of dinitrogen pentoxide (theoretical yield) are formed? (112 g) 2 N2 + O2 N2O5 5 2 LR ER 1 mol N2 5 mol O2 (1) 112 g N2 x x = 9.99 mol O2 28.02 g N2 2 mol N2 LR ER used 1 mol N2 2 mol N2O5 108.02 g N2O5 (2) 112 g N2 x x x 28.02 g 2 mol N2 1 mol N2O5 LR = 432 g N2O5 Theoretical Yield

C. Aqueous sodium chloride and aqueous silver nitrate react forming solid silver chloride (a precipitate) and aqueous sodium nitrate. If one solution contains 100.0 g of silver nitrate and it reacts with unlimited sodium chloride, how many grams of silver chloride (Theoretical Yield) can be made? ER LR NaCl (aq) + AgNO3 (aq) AgCl (s) + NaNO3 (aq) LR 1 mol AgNO3 1 mol AgCl 143.32 g 100.0 g AgNO3 x x x 169.88 g 1 mol AgNO3 1 mol AgCl = 84.37 g AgCl Theoretical Yield

D. How many (1) grams (theoretical yield) and (2) molecules of carbon dioxide gas are produced when 100.0 g of methane gas (CH4) are burned. Methane reacts with oxygen to form carbon dioxide and water. LR CH4 + O2  CO2 + H2O 2 2 ER 1 mol CH4 1 mol CO2 44.01 g CO2 (1) 100.0 g CH4 x x x 16.05 g 1 mol CH4 1 mol CO2 LR = 274.2 g CO2 Theoretical Yield 1 mol CH4 1 mol CO2 6.022 x 1023 molec CO2 (2) 100.0 g CH4 x x x 16.05 g 1 mol CH4 1 mol CO2 LR = 3.752 x 1024 molec CO2

E. Nitrogen reacts with hydrogen gas forming ammonia (nitrogen trihydride). How many (1) grams of hydrogen and (2) moles of nitrogen are required to form 1.00 pound of ammonia? N2 + H2  NH3 3 2 453.6 g NH3 1 mol NH3 3 mol H2 2.02 g H2 (1) 1.00 lb NH3 x x x x 1 lb 17.04 g 2 mol NH3 1 mol H2 = 80.7 g H2 453.6 g NH3 1 mol NH3 1 mol N2 (2) 1.00 lb NH3 x x x = 13.3 mol N2 1 lb 17.04 g 2 mol NH3

Review Question: Nitrogen and hydrogen gas combine to form hydrazine (dinitrogen tetrahydride) which is used for rocket fuel. How many grams of nitrogen gas are needed to form 155 g of hydrazine? N2 + 2 H2  N2H4 x 155 g N2H4 32.06 g 1 mol N2H4 1 mol N2 28.02 g N2 = 135 g N2

F. How much hydrogen peroxide can be formed (theoretical yield) from 10.0 g of hydrogen and 125 g of oxygen? What is the limiting reagent? H2 + O2  H2O2 1 mol H2 1 mol H2O2 34.02 g H2O2 10.0 g H2 x x x = 168 g H2O2 2.02 g 1 mol H2 1 mol H2O2 1 mol O2 1 mol H2O2 34.02 g H2O2 125 g O2 x x x = 133 g H2O2 32.00 g 1 mol O2 1 mol H2O2 L.R. = O2 Theoretical yield

G. Determine the theoretical yield of water that can be formed from 10 G. Determine the theoretical yield of water that can be formed from 10.0 g of hydrogen and 100.0 g of oxygen. What is the limiting reagent? How much excess reagent is remaining? ER 2 H2 + O2  H2O 2 1 mol H2 1 mol H2O 18.02 g H2O 10.0 g H2 x x x = 89.2 g H2O 2.02 g 1 mol H2 1 mol H2O L.R. = H2 Theoretical yield 1 mol O2 2 mol H2O 18.02 g H2O 100.0 g O2 x x x = 112.6 g H2O 32.00 g 1 mol O2 1 mol H2O 1 mol H2 1 mol O2 32.00 g O2 10.0 g H2 x x x = 79.2 g O2 2.02 g 2 mol H2 1 mol O2 ER used 100.0 g O2 79.2 g O2 = 20.8 g O2 remaining

H. How many grams of aluminum oxide can be formed (theoretical yield) from 100.0 g of aluminum and 100.0 g of oxygen gas? How much excess reagent is remaining? 4 Al + O2  Al2O3 3 2 1 mol Al 2 mol Al2O3 101.96 g 100.0 g Al x x x = 189.0 g Al2O3 26.98 g 4 mol Al 1 mol Al2O3 L.R. = Al Theoretical yield 1 mol O2 2 mol Al2O3 101.96 g 100.0 g O2 x x x = 212.4 g Al2O3 32.00 g 3 mol O2 1 mol Al2O3 1 mol Al 3 mol O2 32.00 g O2 100.0 g Al x x x = 88.95 g O2 26.98 g 4 mol Al 1 mol O2 100.0 g O2 - 88.95 g O2 = 11.05 g O2 Remaining

I. A student does a lab in which she makes 2.17 g of Lead (II) nitrate. From her limiting reagent she calculated that she should have made 2.55 g. Determine: (1) theoretical yield, (2) actual yield, (3) % yield (4) absolute and relative error. (1) 2.55 g Pb(NO3)2 (2) 2.17 g Pb(NO3)2 (4) Ea = 2.55 g – 2.17 g = 0.38 g Note: % yield + Er = 100 %

J. A sample of 100. 0 g of hydrazine (N2H4) burns in 280 J. A sample of 100.0 g of hydrazine (N2H4) burns in 280.0 g of oxygen gas and 97.2 g of water are formed. Calculate the % yield of water. (Nitrogen dioxide is also produced.) Calculate the % error (relative error). N2H4 + O2  H2O + NO2 3 2 2 1 mol N2H4 2 mol H2O 18.02 g 100.0 g N2H4 x x x = 112.4 g H2O 32.06 g 1 mol N2H4 1 mol H2O 1 mol O2 2 mol H2O 18.02 g 280.0 g O2 x x x = 105.1 g H2O 32.00 g 3 mol O2 1 mol H2O LR Theo. yield % error = 100% - 92.5% = 7.5%

2 AgNO3 + CaCl2  Ca(NO3)2 + 2 AgCl (s) K. In lab a student mixes a solution that contains 1.70 g silver nitrate and a solution that contains 0.832 g calcium chloride. Aqueous calcium nitrate and solid silver chloride are produced. After washing, filtering, and drying the precipitate, he finds that he has obtained 1.25 g of silver chloride. What is the % yield? 2 AgNO3 + CaCl2  Ca(NO3)2 + 2 AgCl (s) 1.70 g AgNO3 169.88 g 1 mol AgNO3 2 mol AgNO3 2 mol AgCl 1 mol AgCl 143.32 g x = 1.43 g AgCl Theo. yield 0.832 g CaCl2 110.98 g 1 mol CaCl2 2 mol AgCl 1 mol AgCl 143.32 g x = 2.15 g AgCl

IQ1 Determine the mass of Iron III oxide formed when 25.1 g of iron reacts with oxygen. Balanced equations: ______________________________ 2. What mass of oxygen is needed to react with 6.8 x 1023 atoms of iron? (Use the above equations). 4 Fe + 3 O2 → 2 Fe2O3 25.1 g Fe 2 mol Fe2O3 159.70 g Fe2O3 1 mol Fe 55.85 g Fe 4 mol Fe 1 mol Fe2O3 = 35.9 g Fe2O3 6.8 x 1023 atoms Fe 1 mol Fe 3 mol O2 32.00 g O2 6.022 x 1023 atoms Fe 4 mol Fe 1 mol O2 = 27 g O2

IQ 2 Determine the (a) limiting reagent and (b) theoretical yield of iron when 82.3 g of iron (III) oxide reacts with 79.3 g of carbon monoxide. (c) Grams of excess reagent remaining. Balance equ.: __________________________________ (a) (b) Grams excess reagent remaining? Fe2O3 + CO → Fe + CO2 3 2 3 82.3 g Fe2O3 1 mol Fe2O3 3 mol CO 28.01 g CO 159.70 g Fe2O3 1 mol Fe2O3 1 mol CO Fe2O3 – Limiting Reagent 43.3 g CO needed 82.3 g Fe2O3 1 mol Fe2O3 2 mol Fe 55.85 g Fe 159.70 g Fe2O3 1 mol Fe2O3 1 mol Fe 57.6 g Fe Excess = grams @ start – grams reacted Excess = 79.3 g CO – 43.3 g CO = 36.0 g CO

Stoichiometry WS I a. 6.92 mol O2 b. 6.20 g KCl a. 4.07g H2O b. 104 g CO2 a. 0.026 mol FeCl3 b. 15.3 mol Cu

Stoichiometry WS II a. 6.92 mol O2 b. 6.20 g KCl a. 4.07g H2O b. 104 g CO2 a. 0.026 mol FeCl3 b. 15.3 mol Cu

Sample Test Part A (1) Syn (2) Decomp (3) DD (4) SD (5) Combustion a. 1, 4, 3, 4 b. 1, 1, 1, 1 c. 6, 1, 2, 3 d. 2, 3, 1, 3 a. 4, 3, 2 (Syn) b. 2, 2, 2, 1 (SD) c. 2, 7, 4, 6 (Comb) d. 1, 2, 1, 2 (DD) e. 2, 1, 1 (Decomp) a. 1, 1, 2 AgBr, 1 CaSO4 b. 2, 1 H2, 1 Br2 c. 1, 5, 3 CO2, 4 H2O d. 2, 3, 1 Al2S3 e. 2, 1 N2, 3 H2

Sample Test Part A a. 2, 3, 1 Al2(SO4)3 Cu b. 1, 2, 1 Ca(OH)2, 1H2 c. 1, 2, 1 Fe(OH)2, 1 H2 d. 1, 2, 1 ZnCl2, 1 H2 e. NR f. 1, 2, 1 Cu(NO3)2, 2 Ag HOFBrINCl Speeds up rxn w/o being used up in the rxn. C, CO, CO2, H2O See Metal activities series.

Sample Test Part B a. Fe b. O2 a. 1.87 x 1023 molec b. 138 g NaCl c. 6.33 x 1023 molec Cl2 a. AgNO3 b. 69.9 g Ag c. 89 g Cu 74.3 % When LR runs out, the rxn stops. 145 g NH3 F, A , E, C , D, B

Unit 6 HW Packet Schedule Unit 6 Study Guide Stoichiometry WS I Stoichiometry WS II Limiting Reagent & Percent Yield WS Theoretical Yield/Percent Yield WS Reactions WS I Reactions WS II Reactions WS III Stoichiometry Sample Test

Chemistry WS a. 6.92 mol O2 b. 6.20 g KCl a. 4.07g H2O b. 104 g CO2 a. 0.026 mol FeCl3 b. 15.3 mol Cu

Stoichiometry WS 1.81 x 1023 molecules H2 7.3 g H2 3.09 mol Al2(SO4)3 22 g Al 6.7 x 1022 atoms Al 8.1 g H2

Limiting Reagent & Percent Yield WS a. L.R. = HCl b. L.R. = O2 c. L.R. = Cl2 a. L.R. = HCl; 0.26 mol H2 produced b. 0.26 mol Zn used; 0.04 mol Zn excess a. L.R. = HCl; 0.12 mol AlCl3 produced b. 0.12 mol Al used; 0.03 mol Al excess

Limiting Reagent & Percent Yield 87.9 % 75.2 % Theoretical Yield/Percent Yield 54.2 % Actual yield = 1660 g NH3 37.0 % Actual yield = 58.7 g MgO

Reaction Types DD: Al2(SO4)3 + Ca3(PO4)2  2 AlPO4 + 3 CaSO4 DD: MgCl2 + 2 AgNO3  Mg(NO3)2 + 2 AgCl Syn: 2 H2 + O2  2 H2O SD: Zn + Cu(NO3)2  Zn(NO3)2 + Cu DD: CuO + H2SO4  CuSO4 + H2O SD: 2 NaI + Cl2  2 NaCl + I2 Decomp: Cu(OH)2  CuO + H2O 12. SD: K + H2O  KOH + H2

Sample Test Part A (1) Syn (2) Decomp (3) DD (4) SD (5) Combustion a. 1, 4, 3, 4 b. 1, 1, 1, 1 c. 6, 1, 2, 3 d. 2, 3, 1, 3 a. 4, 3, 2 (Syn) b. 2, 2, 2, 1 (SD) c. 2, 7, 4, 6 (Comb) d. 1, 2, 1, 2 (DD) e. 2, 1, 1 (Decomp) a. 1, 1, 2 AgBr, 1 CaSO4 b. 2, 1 H2, 1 Br2 c. 1, 5, 3 CO2, 4 H2O d. 2, 3, 1 Al2S3 e. 2, 1 N2, 3 H2

Sample Test Part A a. 2, 3, 1 Al2(SO4)3 Cu b. 1, 2, 1 Ca(OH)2, 1H2 c. 1, 2, 1 Fe(OH)2, 1 H2 d. 1, 2, 1 ZnCl2, 1 H2 e. NR f. 1, 2, 1 Cu(NO3)2, 2 Ag HOFBrINCl Speeds up rxn w/o being used up in the rxn. C, CO, CO2, H2O See Metal activities series.

Sample Test Part B a. Fe b. O2 a. 1.87 x 1023 molec b. 138 g NaCl c. 6.33 x 1023 molec Cl2 a. AgNO3 b. 69.9 g Ag c. 89 g Cu 74.3 % When LR runs out, the rxn stops. 145 g NH3 F, A , E, C , D, B

Stoichiometry Review WS a. N2O5 + H2O  2 HNO3 b.111.8 g HNO3 c. 15.7 g H2O 2. a. 4 Al + 3 PbO2  2 Al2O3 + 3 Pb b. 2.113 g Al 3. a. LR = H2S b. 8.257 g (NH4)2S c. 2.709 g NH3 4. a. Yes; SD; Cl2 + 2KBr  2 KCl + Br2 b. LR = Cl2 c. 1.93 g Br2 d. 0.335 g KBr 5. a. LR = As2O3 b. 79.3 % 6. a. 6, 6, 1, 6 b. 6.32 x 103 g O2 7. a. 71 % b. 26 metric tons

Unit 6 HW Packet Schedule Unit 6 Study Guide Stoichiometry WS I Stoichiometry WS II Limiting Reagent & Percent Yield WS Theoretical Yield/Percent Yield WS Reactions WS I Reactions WS II Reactions WS III Stoichiometry Sample Test