Unit 1: Stoichiometry Chemistry 2202 1:27 AM

Stoichiometry Stoichiometry deals with quantities used in OR produced by a chemical reaction 1:27 AM

3 Parts Mole Calculations (Chp. 2 & 3) Stoichiometry and Chemical Equations (Chp. 4) Solution Stoichiometry (Chp. 6) 1:27 AM

PART 1 - Mole Calculations Isotopes and Atomic Mass (pp. 43 - 46) Avogadro’s number (pp. 47 – 49) Mole Conversions (pp. 50 - 74) M, MV, NA, n, m, v, N 1:27 AM

Questions p. 54 #’s 5 – 8 p. 45 #’s 1 – 4 p. 65 #’s 2, 4, 5 1:27 AM

PART 1 - Mole Calculations Percent composition: - given mass (p. 79 - 82) - given the chemical formula (p. 83 - 86) Empirical Formulas (pp. 87 - 94) Molecular Formulas (pp. 95 - 98) Lab: Formula of a Hydrate 1:27 AM

Questions p. 103 #’s 23 – 24 p. 82 #’s 1 – 4 p. 86 #’s 1, 3 – 6 1:27 AM

Isotopes and Atomic Mass atomic number - the number of protons in an atom or ion mass number - the sum of the protons and neutrons in an atom isotope - atoms which have the same number of protons and electrons but different numbers of neutrons 1:27 AM

Isotopes and Atomic Mass eg. 1:27 AM

Isotopes and Atomic Mass 1:27 AM

Isotopes and Atomic Mass 11 % 79 % 10 % not all isotopes are created equal 1:27 AM

Isotopes and Atomic Mass 1:27 AM

Isotopes and Atomic Mass atomic mass unit (AMU - p.43) a unit used to describe the mass of individual atoms the symbol for the AMU is u 1 u is 1/12 of the mass of a carbon-12 atom 1:27 AM

Isotopes and Atomic Mass average atomic mass (AAM) the AAM is the weighted average of all the isotopes of an element (p. 45) p. 14 # 5 p. 45 #’s 1 – 4 p. 46 #’s 1 – 6 p. 75 #’s 9 - 12 1:27 AM

Finding % Abundance eg. Br has two naturally occurring isotopes. Br-79 has a mass of 78.92 u and Br-81 has a mass of 80.92 u. If the AAM of Br is 79.90 u, determine the percentage abundance of each isotope. 1:27 AM

Finding % Abundance Let y = fraction of Br-81 x + y = 1 y = 1 - x Let x = fraction of Br-79 Let y = fraction of Br-81 x + y = 1 78.92x + 80.92y = 79.90 y = 1 - x 1:27 AM

x + y = 1 78.92x + 80.92y = 79.90 1:27 AM

Avogadro’s Number (p. 47) p. 48 1:27 AM

Avogadro’s Number The MOLE is a number used by chemists to count atoms The MOLE is the number of atoms contained in exactly 12 g of carbon-12. In honor of Amedeo Avogadro, the number of particles in 1 mol has been called Avogadro’s number. 1:27 AM

How big is Avogadro's number? An Avogadro's number of soft drink cans would cover the surface of the earth to a depth of over 200 miles. Avogadro's number of unpopped popcorn kernels spread across the USA, would cover the entire country to a depth of over 9 miles. 1:27 AM

How big is Avogadro's number? If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole. 1:27 AM

Avogadro’s Number NA = 6.022 x 1023 particles/mol 1 mole = 6.02214199 x 1023 particles 1 mol = 6.022 x 1023 particles NA = 6.022 x 1023 particles/mol 1:27 AM

Avogadro’s Number 1:27 AM

Avogadro’s Number 5 mol 3.011 x 1024 atoms 0.01 mol 6.022 x 1021 atoms Number of moles Number of atoms 5 mol 3.011 x 1024 atoms 0.01 mol 6.022 x 1021 atoms 7.72 mol 4.65 x 1024 atoms 0.0133 mol 8.01 x 1021 atoms 1:27 AM

N = n x NA Avogadro’s Number Formulas: n = # of moles N = # of particles (atoms, ions, molecules, or formula units) NA = Avogadro’s # 1:27 AM

Avogadro’s Number How many moles are contained in the following? 2.56 x 1028 Pb atoms 7.19 x 1021 CO2 molecules 1:27 AM

eg. Calculate the number of moles in 4.98 x 1025 atoms of Al. Avogadro’s Number eg. Calculate the number of moles in 4.98 x 1025 atoms of Al. eg. How many formula units of Na2SO4 are in 5.69 mol of Na2SO4? # of Na ions? # of Oxygen atoms? 1:27 AM

Avogadro’s Number 1. How many molecules of glucose are in 0.435 mol of C6H12O6? How many carbon atoms? 2. Calculate the number of moles in a sample of glucose that has 3.56 x 1022 hydrogen atoms. 1:27 AM

Avogadro’s Number pp. 51 – 53: #’s 5 – 15 p. 54:  #’s 4 - 8 1:27 AM

Molar Mass The mass of one mole of a substance is called the molar mass of the substance eg. 1 mole of Pb has a mass of 207.19 g 1 mole of Ag has a mass of 107.87 g 1:27 AM

Molar Mass The symbol for molar mass is M and the unit is g/mol eg. MPb = 207.19 g/mol MAg = 107.87 g/mol 1:27 AM

Molar Mass The molar mass of a compound is the sum of the molar masses of the elements in the compound eg. Calculate the molar mass of: a) H2O b) C6H12O6 c) Ca(OH)2 1:27 AM

Molar Mass H2O has 2 hydrogens and 1 oxygen 2 x 1.01 = 2.02 18.02 g/mol 1:27 AM

Molar Mass C6H12O6 6 x 12.01 = 72.06 12 x 1.01 = 12.12 6 x 16.00 = 96.00 180.18 g/mol 1:27 AM

Molar Mass Your calculator may not show Ca(OH)2 the zeroes. There should be 2 digits after the decimal when adding molar masses Ca(OH)2 1 x 40.08 = 40.08 2 x 16.00 = 32.00 2 x 1.01 = 2.02 74.10 g/mol 1:27 AM

Molar Mass p. 57: #’s 16 – 19 & Molar Masses Handout 1. 151.92 g/mol 7. 58.44 g/mol 2. 120.38 g/mol 8. 100.09 g/mol 3. 105.99 g/mol 9. 44.02 g/mol 4. 100.40 g/mol 10. 248.22 g/mol 5. 74.44 g/mol 11. 115.04 g/mol 6. 78.01 g/mol 1:27 AM

Molar Mass Calculations Avogadro’s # molar mass N = n x NA m = n x M 1:27 AM

Molar Mass Calculations m = n x M N = n x NA 1:27 AM

Molar Mass Calculations eg. How many moles are in 25.3 g of NO2? m = 25.3 g MNO2 = 46.01 g/mol 1:27 AM

Molar Mass Calculations eg. What is the mass of 4.69 mol of water? n = 4.69 mol Mwater = 18.02 g/mol m = n x M = (4.69 mol)(18.02 g/mol) = 84.5 g 1:27 AM

Molar Mass Calculations Practice: p. 59 #’s 20 - 23 p. 60 #’s 24 - 27 1:27 AM

The Mole #4 answers 1.a) 0.038 mol 2.a) 17.4 g 3.a) 5631 g b) 3.75 mol b) 1560. g b) 3.73 g c) 0.276 mol c) 528 g c) 10.02 g d) 23 mol d) 97513 g d) 1662.9 g e) 0.0575 mol e) 9328 g e) 981.2 g 4.a) 1.82 mol c) 0.02133 mol e) 0.0000573 mol b) 13 mol d) 4.3423 mol Molar Masses 1.a) 207.19 b) 12.01 c) 18.02 d) 149.12 e)245.30 2.a) 6.94 b) 44.01 c) 32.00 d) 78.01 e) 141.94 3.a) 63.55 b) 107.87 c) 270.87 d) 393.13 e) 327.49 4.a) 6.94 b) 196.97 c) 109.93 d) 136.18 e) 64.07 1:27 AM

Particle–Mole-Mass Conversions eg. How many molecules are in 26.9 g of water? m = 26.9 g Mwater= 18.02 g/mol NA = 6.022 x 1023 molecules/mol Find N 1:27 AM

Particle–Mole-Mass Conversions N = n x NA = 1.493 X 6.022 x 1023 = 8.99 x 1023 molecules = 1.493 mol H2O 1:27 AM

Particle–Mole-Mass Conversions eg. A sample of Sn contains 4.69 x 1028 atoms. Calculate its mass. N = 4.69 x 1028 NA = 6.022 x 1023 molecules/mol MSn = 118.69 g/mol Find m 1:27 AM

Particle–Mole-Mass Conversions m = n x M = 77881 mol x 118.69 g/mol = 9.24 x 10 6 g = 77,881 mol 1:27 AM

Particle–Mole-Mass Conversions Calculate the mass of 4.80 x 1024 water molecules. How many grams are in 2.53 x 1026 CH4 molecules? How many atoms are in 68.0 g of Sn? Calculate the number of molecules in 105 g of C6H12O6. 1:27 AM

Particle–Mole-Mass Conversions Particles (N) Moles (n) Mass (m) 4.80 x 1024 H2O molecules 2.53 x 1026 CH4 molecules 68.0 g of Sn 105 g of C6H12O6 1:27 AM

Particle–Mole-Mass Conversions particles (N) Moles (n) Mass (m) 5.98 x 1026 Cu atoms 4.50 g H2O 6.15 mol O3 1:27 AM

Particle–Mole-Mass Conversions particles (N) Moles (n) Mass (m) 4.18 x 1022 Al atoms 2.45 g C8H18 0.145 mol S2F4 1:27 AM

Particle–Mole-Mass Conversions x NA x M particles(N) moles (n) mass (m) ÷ M ÷ NA Practice: p. 63 #’s 28 - 33 p. 64 #’s 34 – 37 p. 76 # 15 1:27 AM

Particle–Mole-Mass Conversions eg. How many molecules are in 4.78 g of glucose? m = 4.78 g Mwater= 180.18 g/mol NA = 6.022 x 1023 molecules/mol Find N 1:27 AM 1:27 AM

Particle–Mole-Mass Conversions N = n x NA = 0.02653 X 6.022 x 1023 = 8.99 x 1023 molecules = 0.02653 mol glucose 1:27 AM 1:27 AM

Particle–Mole-Mass Conversions Practice: p. 63 #’s 28 - 33 p. 64 #’s 34 – 37 p. 76 # 15 1:27 AM

Molar Mass Calculations Practice: p. 54 #’s 5 - 8 p. 65 #’s 2, 4, 5 p. 75 #’s 13, 14, p. 76 #’s 15, 17 – 19, 21 -23 1:27 AM

Molar Volume The volume of a gas increases when temperature increases but decreases when pressure increases . The volume of gases is measured under conditions of Standard Temperature and Pressure (STP) 1:27 AM

Molar Volume Standard Pressure – 101.3 kPa Standard Temperature – 0 °C Avogadro hypothesized that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. 1:27 AM

Molar Volume Experimental evidence shows the volume of one mole of ANY GAS at STP is 22.4 L/mol OR VSTP = 22.4 L/mol 1:27 AM

Molar Mass Calculations given volume in Litres Molar Mass Calculations N = n x NA v = n x VSTP m = n x M 1:27 AM

Molar Mass Calculations x NA x M particles(N) moles (n) mass (m) ÷ M ÷ NA x VSTP ÷ VSTP volume (v) 1:27 AM 1:27 AM

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Molar Volume p. 73 #’s 38 – 43 p. 74 #’s 1 – 4 p. 76 #’s 26, 27 WorkSheet: The Mole #6 1:27 AM

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Percent Composition (p. 79) The mass percent of a compound is the mass of an element in a compound expressed as a percent of the total mass of the compound. 1:27 AM

Percent Composition eg. 8.50 g of a compound was analyzed and found to contain 6.00 g of hydrogen and 2.50 g of carbon. Calculate the mass percent for each element. p. 82 #’s 1 - 4 1:27 AM

Percent Composition mass percent may be found using the formula & the molar mass of a compound. eg. Find the percentage composition for CH4 1:27 AM

Percent Composition p. 85 #’s 5 - 8 M = 12.01 g/mol + 4(1.01 g/mol) 1:27 AM

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p. 85 #’s 5 - 8 p. 86 #’s 1, 3 – 6 p. 107 #’s 6 – 10 1:27 AM

Empirical Formulas An empirical formula gives the simplest ratio of elements in a compound. A molecular formula shows the actual number of atoms in a molecule of a compound. Ionic compounds are always written as empirical formulas 1:27 AM

Empirical Formulas Compound Molecular Formula Empirical Formula butane C4H10 glucose C6H12O6 water H2O benzene C6H6 C2H5 CH2O H2O CH 1:27 AM

Empirical Formulas (p.87) 1:27 AM

Empirical Formulas The empirical formula of a compound may be determined by using the % composition of a given compound. 1:27 AM

Empirical Formulas Method: assume you have 100.0 g of the compound (ie. change % to g) calculate the moles (n) for each element divide each n by the smallest n to get the ratio for the empirical formula 1:27 AM

Empirical Formulas eg. A compound was analyzed and found to contain 87.4% N and 12.6 % H by mass. Determine the empirical formula of the compound. p. 89 #’s 9 - 12 1:27 AM

Empirical Formulas When finding the EF, the mole ratio may not be a whole number ratio. eg. A compound contains 84.73% N and 15.27 % H by mass. Determine the empirical formula of the compound. 1:27 AM

p. 90 1:27 AM

Empirical Formulas eg. A compound contains 89.91% C and 10.08 % H by mass. Determine the empirical formula of the compound. p. 91 #’s 13 – 16 p. 94 #’s 2-4, 6, 7 Answers on p. 109 1:27 AM

MgO Lab 1:27 AM

Molecular Formulas The molecular formula of a compound is a multiple of the empirical formula. See p. 95 1:27 AM

Molecular Formulas To find the molecular formula we need the empirical formula and the molar mass of the compound eg. The empirical formula of hydrazine is NH2. The molar mass of hydrazine is 32.06 g/mol. What is the molecular formula for hydrazine? 1:27 AM

Molecular Formulas p. 97 #’s 17 – 20 p. 107, 108 #’s 11 - 14 1:27 AM

CHC analyzer (p. 99 – 101) 1. Describe the operation of a carbon hydrogen combustion analyzer. 2. 22.0 g of carbon dioxide and 10.8 g of water is collected in a CHC analysis. Determine the empirical formula of the hydrocarbon. p. 101 #’s 21, 22 1:27 AM

Formula of a hydrate To determine the formula of a hydrate: - calculate the moles of water - calculate the moles of anhydrous compound - determine the simplest ratio 1:27 AM

Formula of a hydrate eg. Use the data below to determine the value of x in LiCl• xH2O. mass of crucible = 26.35 g crucible + hydrate = 42.15 g crucible + anhydrous compound= 34.94 g 1:27 AM

mwater = 42.15 – 34.94 = 7.21 g H2O mLiCl = 34.94 – 26.35 = 8.59 g LiCl 1:27 AM

eg. Na2CO3. xH2O crucible = 15.96 g crucible + hydrate = 22.19 g crucible + anhydrous compound = 19.67 g 1:27 AM

crucible + anhydrous compound = 158.23 g p. 103; # 24 Lab: pp. 104-105 eg. CoCl2.xH2O crucible = 151.96 g crucible + hydrate = 164.35 g crucible + anhydrous compound = 158.23 g p. 103; # 24 Lab: pp. 104-105 1:27 AM

Formula of a hydrate mass of empty beaker mass of beaker & hydrate mass of beaker & anhydrous compound 1:27 AM

Review – Chp. 3 p. 86 #’s 1, 3 – 6 p. 94 #’s 1 – 7 p. 103 # 23 1:27 AM

Test p. 54 #’s 5 – 8 p. 45 #’s 1 – 4 p. 65 #’s 2, 4, 5 p. 46 #’s 1 – 6 1:27 AM

Test p. 82 #’s 1 – 4 p. 85 #’s 5 – 8 p. 89 #’s 9 – 12 p. 91 #’s 13 – 16 p. 97 #’s 17 - 20 p. 103 #’s 23 – 24 p. 86 #’s 1, 3 – 6 p. 94 #’s 1 - 7 p. 106 #’s 1 - 3, 6, 7 p. 107 – 109 #’s 5 – 23, 25 1:27 AM

Test Friday – Dec. 4 1. Formula of a hydrate - lab activity - p. 103 # 24 % composition - given mass p. 82 #’s 1 – 4 - given formula p. 85 #’s 5 - 8 p. 86 #’s 3 & 4 p. 103 # 23 p. 107 # 5 EF and MF - What is an empirical formula - Finding EF from % composition pp. 89, 91 #’s 9 - 16 (watch for .5 or .333) - Finding MF from EF and molar mass p. 97 #’s 17 – 20 p. 108 #13 4. Mole calculations – Chp. 2 1:27 AM

Stoichiometry (Chp.4) Stoichiometry is the determination of quantities needed for, or produced by, chemical reactions. Ratios from balanced chemical equations are used to predict quantities. 1:27 AM

Clubhouse sandwich recipe Stoichiometry – p. 111 Clubhouse sandwich recipe 1:27 AM

Clubhouse sandwich recipe Fill in the missing quantities: Slices of Toast Slices of Turkey Strips of Bacon # of Sandwiches 12 27 66 100

Mole Ratios A mole ratio is a mathematical expression that shows the relative amounts of two species involved in a chemical change. 1:27 AM

Mole Ratios A mole ratio Comes from a balanced chemical equation Shows the relative amounts of the reactants/products in moles Looks like 1:27 AM

N2(g) + 3 H2 (g) → 2 NH3 (g) 20 66 140 81 1:27 AM 1:27 AM

C3H8 + 5 O2 → 3 CO2 + 4 H2O How many moles of CO2 are produced when 31.5 mol of O2 react? 1:27 AM

C3H8 + 5 O2 → 3 CO2 + 4 H2O How many moles of H2O are produced when 1.35 mol of O2 react? 1:27 AM 1:27 AM

C3H8 + 5 O2 → 3 CO2 + 4 H2O How many moles of C3H8 are needed to react with 0.369 mol of O2? 1:27 AM 1:27 AM

C5H12 + O2 → CO2 + H2O How many moles of CO2 are produced when 6.35 mol of O2 react? 1:27 AM 1:27 AM

Al(s) + Br2(l) → AlBr3(s) How many moles of Br2 are needed to produce 0.315 mol of AlBr3? p. 115 #’s 4 – 7 p. 117 #’s 8 - 10 1:27 AM 1:27 AM

Mole to Mole Stoichiometry How many moles of copper would be produced if 20.5 mol of copper (II) oxide decomposes? 1:27 AM 1:27 AM

Mass to Mole Stoichiometry How many moles of water are produced when 20.6 g of CH4 burns? 1:27 AM 1:27 AM

Mass to Mole Stoichiometry How many moles of nitrogen gas are needed to produce 6.75 g of NH3 in a reaction with hydrogen gas? 1:27 AM

Mass to Mole Stoichiometry How many moles of silver would be produced if 10.0 g of silver nitrate reacts with copper metal? 1:27 AM

Mass to Mole Stoichiometry How many moles of water are produced when 20.6 g of C6H12 burns? 1:27 AM

Mass to Mole Stoichiometry How many moles of water are produced when 20.6 g of CH4 burns? 1:27 AM 1:27 AM

Mass to Mole Stoichiometry 1:27 AM 1:27 AM

Mole to Mass Stoichiometry What mass of CaCl2 is produced when 4.38 mol of Ca(NO3)2 reacts with NaCl? 1:27 AM

Mole to Mass Stoichiometry What mass of copper would be produced if 20.5 mol of CuO decomposes? 1:27 AM 1:27 AM

Four step stoichiometry Write a balanced chemical equation Calculate moles given Mole ratio – find moles required Calculate required quantity m = n x M v = n x Vstp N = n x NA 1:27 AM

Mass to Mass Stoichiometry How many grams of water are produced when 5.45 g of C3H8 burns? 1:27 AM 1:27 AM

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Mass to Mass Stoichiometry eg. Calculate the mass of HCl needed to react with 3.56 g of Fe to produce FeCl2. 1:27 AM 1:27 AM

Stoichiometry (Chp.4) eg. What mass of CO2 gas is produced when 45.9 g of CH4 burns ? Step #1 CH4 + 2 O2 → CO2 + 2 H2O 45.9 g ? g 1:27 AM

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eg. How many moles of HCl needed to react with 3 eg. How many moles of HCl needed to react with 3.56 g of Fe to produce FeCl2. 1:27 AM

Mole Calculations (p. 121 #13) 3.56 g ? g Fe + 2 HCl → FeCl2 + H2 Step #2 = 0.06374 mol Fe Step #3 = 0.12748 mol HCl 1:27 AM

Mole Calculations p. 122 #15 Given 32.0 g of sulfur (M = 256.56 g/mol) Find mass of ZnS #2 n = 0.1247 mol S8 #3 n = 0.9976 mol ZnS (M = 97.45 g/mol) #4 m = 97.2 g ZnS 1:27 AM

Mole Calculations p. 123 #18 Given 33.5 g of H3PO4 (M = 98.00 g/mol) Find mass of MgO #2 n = 0.3418 mol H3PO4 #3 n = 0.5128 mol MgO (M = 40.31 g/mol) #4 m = 20.7 g MgO 1:27 AM

Mole Calculations p. 123 #17 Given 25.0 g of Al4C3 (M = 143.95 g/mol) Find volume of CH4 #2 n = 0.174 mol Al4C3 #3 n = 0.522 mol CH4 (MV = 22.4 L/mol) #4 m = 11.7 L CH4 1:27 AM 1:27 AM

How many moles of aluminum chloride can be produced from the reaction of chlorine and 10.8 mol of aluminum ? Cl2(g) + Al(s) → AlCl3(s) 1:27 AM

How many moles of magnesium are needed to react with 27 g of iodine to form magnesium iodide? 1:27 AM

How many grams of nitrogen are needed to react with 14 How many grams of nitrogen are needed to react with 14.0 mol of oxygen to produce nitrogen dioxide ? N2(g) + O2(g) → NO2(g) 1:27 AM

Mole Calculations (p. 121 #14) 2.34 g ? L #2 n = 0.05086 mol NO2 #3 n = 0.01272 mol O2 (MV = 22.4 L/mol) #4 n = (0.01272)(22.4) = 0.285 L O2 1:27 AM

Limiting Reactant (p. 128) 10.0 g of Li reacts with 15.0 g of Br2. Calculate the mass of LiBr produced. Half the class finds yield from 10.0 g of Li. Other half find yield from 15.0 g of bromine. 1:27 AM

Limiting Reactant (p. 128) 2 Li Br2 2 LiBr 10.0 15.0 1:27 AM WorkSheet: LR Introduction (wp file) 1:27 AM 131

Limiting Reactant (p. 128) The Limiting Reactant (LR) OR Limiting Reagent (LR) is the substance that is completely used in a chemical reaction. The Excess Reactant is the reactant that is left over after a reaction is complete. 1:27 AM

Limiting Reactant (p. 128) eg. 2.00 g of NaI reacts with 2.00 g of Pb(NO3)2. Determine the LR and calculate the amount of PbI2 produced. write a balanced equation find n for each reactant (Step #2) find moles produced by each reactant (Step #3) 1:27 AM

Pb(NO3)2 + 2 NaI → 2 NaNO3 + PbI2 2.00 g 2.00 g nPb(NO)3 = 2.00 g 331.21 g/mol = 0.006038 mol nNaI = 2.00 g 149.89 g/mol = 0.013343 mol 1:27 AM

nPbI2 = 0.006038 mol Pb(NO3)2 x 1 mol PbI2 1 mol Pb(NO3)2 nPbI2 = 0.013343 mol NaI x 1 mol PbI2 2 mol NaI = 0.006672 mol PbI2 mPbI2 = 0.006038 mol x 460.99 g/mol = 2.78 g PbI2 1:27 AM

What mass of calcium carbonate will be produced when 20 What mass of calcium carbonate will be produced when 20.0 g of calcium phosphate reacts with 15.0 g of sodium carbonate? (14.2 g) 3 Na2CO3 + Ca3(PO4)2 → 3 CaCO3 + 2 Na3PO4 1:27 AM

Ba(NO3)2 + NaOH → Ba(OH)2 + NaNO3 What mass of barium hydroxide will be produced when 10.0 g of barium nitrate reacts with 30.0 g of sodium hydroxide? (6.56 g) Ba(NO3)2 + NaOH → Ba(OH)2 + NaNO3 = 0.03826 mol Ba(NO3)2 = 0.750 mol NaOH 1:27 AM

Using Ba(NO3)2 = 0.03826 mol Ba(OH)2 Using NaOH = 0.375 mol Ba(OH)2 1:27 AM

What volume of hydrogen gas at STP will be produced when 10 What volume of hydrogen gas at STP will be produced when 10.0 g of zinc metal reacts with 20.0 g of hydrogen chloride? Zn + 2 HCl → H2 + ZnCl2 1:27 AM

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Law of Conservation of Mass (p. 118) In a chemical reaction, the total mass of reactants always equals the total mass of products. eg. 2 Na3N → 6 Na + N2 When 500.00 g of Na3N decomposes 323.20 g of N2 is produced. How much Na is produced in this decomposition?

Law of Conservation of Mass ( p. 118) eg. To produce 90.1 g of water, what mass of hydrogen gas is needed to react with 80.0 g of oxygen? eg. If 3.55 g of chlorine reacts with exactly 2.29 g of sodium, what mass of NaCl will be produced? 1:27 AM

Percent yield (p. 137) The theoretical yield is the amount of product that we calculate using stoichiometry The actual yield is the amount of product obtained from a chemical reaction 1:27 AM

Percent yield (p. 137) p. 139 #’s 31, 32, & 33 1:27 AM

DEMO: silver nitrate + copper Equation: Mass AgNO3 = Mass Cu = 1:27 AM

DEMO: silver nitrate + copper Mass of filter paper and precipitate = Mass of empty filter paper = Mass of precipitate = 1:27 AM