Chapter 2 Overview! all matter, whether liquid, solid, or gas,consists of atoms, which form molecules! the identity of an atom and its chemical behavior is strictly defined by the number of electrons in the outer electronic sphere, which equals to the number of protons in the nucleus! the atomic weight is the combined mass of protons and neutrons in the nucleus. Each contributes one unit of mass. The mass of the electrons is negligible Chapter 2 Overview 6 electrons 6 electrons 6 n o 6 p + 7 n o 6 p + 12 C 13 C number of neutrons ISOTOPES OF CARBON N = Mass - Z atomic number number of protons
Periodic Table Main Group Elements:Group Number corresponds to ionic charge Chapter 2 Overview Ionic compounds: Simple salts. Charges Balanced NaCl Na + Cl LiI Li + I Cs 2 SO 4 Cs + SO 2 4 Fe 2 (CO 3 ) 3 Fe 3+ CO 2 3 (NH 4 ) 3 PO 4 NH + 4 PO 3 4
Chapter 2 Overview Molecular Weight = Sum of Atomic Weights MW(NaCl) = Atomic Mass of Na + At. Mass of Cl Chapter 2 Sample Problem 2.132. Succinic acid is an important metabolite in biological energy production. Give the molecular formula, empirical formula, and molecular mass of succinic acid. H O H H O C C C C O H H O succinic acid H
Chapter 2 Sample Problem 2.132. Succinic acid is an important metabolite in biological energy production. Give the molecular formula, empirical formula, and molecular mass of succinic acid. mol. formula:count all atoms: C 4 H 6 O 4 H O H H O C C C C O H H O succinic acid H Chapter 2 Sample Problem 2.132. Succinic acid is an important metabolite in biological energy production. Give the molecular formula, empirical formula, and molecular mass of succinic acid. mol. formula:count all atoms: C 4 H 6 O 4 empirical formula: C 2 H 3 O 2 (whole numbers!) H O H H O C C C C O H H O succinic acid H
Chapter 2 Sample Problem 2.132. Succinic acid is an important metabolite in biological energy production. Give the molecular formula, empirical formula, and molecular mass of succinic acid. mol. formula:count all atoms: C 4 H 6 O 4 empirical formula: C 2 H 3 O 2 (whole numbers!) H O H H O C C C C O H H O succinic acid H MW = 4!12.01 + 6!1.008 + 4!16.00 = 118.09 Chapter 2 Sample Problem 2.128. How can iodine (Z = 53) have a higher atomic number yet lower atomic weight than tellurium (Z = 52)?
Stoichiometry: The Mole describes quantitative aspects of chemical reactions One mole (1 mol) is 6.022!10 23 molecules (ions, atoms, etc.) Avogadro s number M = MW molar weight (g/mol) molecular weight Stoichiometry MW(H 2 ) = 2.02 MW(NaCl) = 58.44 MW(AgNO 3 ) = 169.91 M = 2.02 g/mol M = 58.44 g/mol M = 169.91 g/mol 1 : 1 ratio of NaCl and AgNO 3 : 58.44 g of NaCl and 169.91 g of AgNO 3 or 0.584 g of NaCl and 0.169 g of AgNO 3 etc
Stoichiometry mass in grams molar weight (g/mol) M (g/mol) = m (g) n (mol) number of mols Stoichiometry mass in grams molar weight (g/mol) M (g/mol) = m (g) n (mol) number of mols n (mol) = m (g) M (g/mol)
Stoichiometry mass in grams m (g) = n (mol)! M (g/mol) number of moles (mol) molar weight (g/mol) Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate?
Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate? divide given mass by mol weight Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate? divide given mass by mol weight what is ammonium carbonate? n (mol) = m (g) M (g/mol)
Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate? m (g) n (mol) = divide given mass by mol weight M (g/mol) what is ammonium carbonate? NH + 4,CO 2 3 " (NH 4 ) 2 CO 3 what is its mol. weight? Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate? m (g) n (mol) = divide given mass by mol weight M (g/mol) what is ammonium carbonate? NH + 4,CO 2 3 " (NH 4 ) 2 CO 3 what is its mol. weight? 2!14.01 + 8!1.008 + 12.01 + 3!16.00 = 96.09 g/mol
Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate? n (mol) = divide given mass by mol weight what is ammonium carbonate? NH + 4,CO 2 3 " (NH 4 ) 2 CO 3 what is its mol. weight? 2!14.01 + 8!1.008 + 12.01 + 3!16.00 = m (g) M (g/mol) 96.09 g/mol answer: 41.6 g : 96.06 g/mol = 0.433 mol Sample Problem 3.2 Ammonium carbonate, is a white solid that decomposes with warming. Among its many uses, it is a component of baking powder, fire extinguishers, and smelling salts. How many moles are in 41.6 g of ammonium carbonate? n (mol) = divide given mass by mol weight what is ammonium carbonate? NH + 4,CO 2 3 " (NH 4 ) 2 CO 3 what is its mol. weight? 2!14.01 + 8!1.008 + 12.01 + 3!16.00 = m (g) M (g/mol) 96.09 g/mol answer: 41.6 g : 96.06 g/mol = 0.433 mol (number of formula units = 0.433 mol! 6.022!10 23 mol -1 = 2.61!10 23 )
Mass Percent from Chemical Formula Glucose (C 6 H 12 O 6 ) is the most important nutrient in the living cell for generating energy. What is the mass percent of each element in glucose? let s focus on carbon first Mass Percent from Chemical Formula Glucose (C 6 H 12 O 6 ) is the most important nutrient in the living cell for generating energy. What is the mass percent of each element in glucose? let s focus on carbon first mass of C in one mole(cule). mass of C 6 H 12 O 6 in one mole(cule)
Mass Percent from Chemical Formula Glucose (C 6 H 12 O 6 ) is the most important nutrient in the living cell for generating energy. What is the mass percent of each element in glucose? let s focus on carbon first mass of C in one mole(cule). mass of C 6 H 12 O 6 in one mole(cule) 6!12.01. = 72.06. = 0.400 = 40.0 mass% 6!12.01 + 12!1.008 + 6!16.00 180.156 Elemental Analysis and Molecular Formula complete burning, than analysis of oxidation products used to determine empirical formula of an unknown compound (gives an empirical formula only!) used to confirm the structure (mass spectroscopy is a superior modern technique that gives a molecular formula directly)
Elemental Analysis and Molecular Formula Sample Problem 3.5: During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) determine the empirical formula of lactic acid (b) determine the molecular formula of lactic acid Elemental Analysis and Molecular Formula Sample Problem 3.5: During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) determine the empirical formula of lactic acid (b) determine the molecular formula of lactic acid mass% # arbitrary weight (let s take 100 g) convert grams to moles
Elemental Analysis and Molecular Formula Sample Problem 3.5: During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) determine the empirical formula of lactic acid (b) determine the molecular formula of lactic acid mass% # arbitrary weight (let s take 100 g) convert grams to moles 40.0 g C = 12.01 g/mol = 3.33 mol H = 6.71 g = 6.66 mol 1.008 g/mol 53.3 g O = = 3.33 mol 16.00 g/mol Elemental Analysis and Molecular Formula Sample Problem 3.5: During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) determine the empirical formula of lactic acid (b) determine the molecular formula of lactic acid mass% # arbitrary weight (let s take 100 g) convert grams to moles 40.0 g C = 12.01 g/mol = 3.33 mol H = 6.71 g = 6.66 mol 1.008 g/mol 53.3 g O = 16.00 g/mol = 3.33 mol CH 2 O
Elemental Analysis and Molecular Formula Sample Problem 3.5: During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0 mass% C, 6.71 mass% H, and 53.3 mass% O. (a) determine the empirical formula of lactic acid (b) determine the molecular formula of lactic acid Empirical formula: CH 2 O " M = 30.03 g/mol Actual M is 3 (three) times greater, therefore: Molecular formula: C 3 H 6 O 3 Chemical Equations Typical Structure: Starting Material(s) # Products(s) H 2 + Cl 2 # 2HCl H H + = Cl Cl Cl H Cl H
Chemical Equations n (mol) = H 2 + Cl 2 # 2HCl m (g) M (g/mol) 0.124 mol H 2 0.124 mol Cl 2 2!0.124 mol HCl 23.12 mol H 2 23.12 mol Cl 2 2!23.12 mol HCl 0.905 mol H 2 0.745 mol Cl 2 Chemical Equations n (mol) = H 2 + Cl 2 # 2HCl m (g) M (g/mol) 0.124 mol H 2 0.124 mol Cl 2 2!0.124 mol HCl 23.12 mol H 2 23.12 mol Cl 2 2!23.12 mol HCl 0.905 mol H 2 0.745 mol Cl 2 2!0.745 mol HCl
Chemical Equations n (mol) = H 2 + Cl 2 # 2HCl m (g) M (g/mol) Global Strategy: Input (g, ml, kg, etc) " moles (starting) " moles (final) " Output (g, ml, kg, etc) Balancing Chemical Equations Starting Material(s) # Products(s) space for coefficients! the same number of atoms of each element on both sides
Balancing Chemical Equations Unbalanced equation: PF 3 + HCl # PCl 3 + HF Balancing Chemical Equations Unbalanced equation: PF 3 + HCl # PCl 3 + HF PF 3 + HCl # PCl 3 + 3HF
Balancing Chemical Equations Unbalanced equation: PF 3 + HCl # PCl 3 + HF PF 3 + HCl # PCl 3 + 3HF PF 3 + 3HCl # PCl 3 + 3HF Balancing Chemical Equations Unbalanced equation: PF 3 + HCl # PCl 3 + HF PF 3 + HCl # PCl 3 + 3HF Balanced equation!: PF 3 + 3HCl # PCl 3 + 3HF Check
Balancing Chemical Equations Unbalanced equation: C 8 H 18 + O 2 # CO 2 + H 2 O Balancing Chemical Equations Unbalanced equation: C 8 H 18 + O 2 # CO 2 + H 2 O Balanced equation: C 8 H 18 + O 2 # 8CO 2 + H 2 O C 8 H 18 + O 2 # 8CO 2 + 9H 2 O C 8 H 18 + 12.5O 2 # 8CO 2 + 9H 2 O 16+9=25 2C 8 H 18 + 25O 2 # 16CO 2 + 18H 2 O Check
Stoichiometric Equivalents C 3 H 8 + 5O 2 # 3CO 2 + 4H 2 O (propane) 1 mol of C 3 H 8 is equivalent to 4 mol of H 2 O 3 mol of CO 2 is equivalent to 5 mol of O 2 3 mol of CO 2 is equivalent to 4 mol of H 2 O Stoichiometric Equivalents C 3 H 8 + 5O 2 # 3CO 2 + 4H 2 O In the combustion of propane, how many moles of carbon dioxide are produced along with 10.0 moles of water? 10 4 = 2.5 2.5!3 = 7.5 " 7.5 moles of CO 2
Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = 180.16) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l).
Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = 180.16) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l). Fundamental Strategy: moles (initial) " moles (final) " Input (g, ml, kg, etc) " moles (starting) " moles (final) " Output (g, ml, kg, etc) Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = 180.16) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l). 1 m 3 = 1000 L, 1kg = 1000 g, d = 1.34 kg/m 3 d = m V
Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = 180.16) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l). 1 m 3 = 1000 L, 1kg = 1000 g, d = 1.34 kg/m 3 d = m V m(o 2 ) = 25 m 3!1.34 kg/m 3 = 33.5 kg = 33500 g n = m M n(o 2 ) = 33500 g 32.00 g/mol = 1047 mol Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = 180.16) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l). n(o 2 ) = 1047 mol n(c 6 H 12 O 6 ) = (1/6)!n(O 2 ) = (1/6)!1047 mol = 174.5 mol n = m M
Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = 180.16) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l). n(o 2 ) = 1047 mol n(c 6 H 12 O 6 ) = (1/6)!n(O 2 ) = (1/6)!1047 mol = 174.5 mol n = m M m(c 6 H 12 O 6 ) = 174.5 mol! 180.16 g/mol = 31438 g Stoichiometric Equivalents 6CO 2 + 6H 2 O # C 6 H 12 O 6 + 6O 2 glucose (MW = 180.16) How many grams of glucose are produced during photosynthesis that generates 25 m 3 of oxygen (at normal conditions, d = 1.34 g/l). n(o 2 ) = 1047 mol n(c 6 H 12 O 6 ) = (1/6)!n(O 2 ) = (1/6)!1047 mol = 174.5 mol n = m M m(c 6 H 12 O 6 ) = 174.5 mol! 180.16 g/mol = 31438 g
Example 2H 2 + O 2 # 2H 2 O 16.0 g 16.0 g? g n (mol) = m (g) M (g/mol) Example 2H 2 + O 2 # 2H 2 O 16.0 g 16.0 g? g 7.94 mol 0.5 mol n (mol) = m (g) M (g/mol)
Example 2H 2 + O 2 # 2H 2 O 16.0 g 16.0 g? g 7.94 mol 0.5 mol 1.0 mol n (mol) = m (g) M (g/mol) Example 2H 2 + O 2 # 2H 2 O 16.0 g 16.0 g? g 7.94 mol 0.5 mol 1.0 mol " 18.02 g n (mol) = m (g) M (g/mol)
Yields: Theoretical, Percent 2A + B # A 2 B + c + d +... 0.343 g should give 0.772 g actually, 0.645 g was obtained Yield = Y = 0.645 g 0.772 g!100% = 84% n (mol) = m (g) M (g/mol) Yields: Theoretical, Percent 2A + B # A 2 B + c + d +... 0.200 mol should give 0.100 mol actually, 0.084 mol was obtained Yield = Y = 0.084 mol 0.100 mol!100% = 84% n (mol) = m (g) M (g/mol)
Yields: Overall 90% 75% 85% A # B # C # D Overall Y = 0.90! 0.75! 0.85! 100% = 57% 57% A # D Solutions Solute Solvent concentration: a quantitative description of a solution
Solutions Molarity = moles of solute liters of solution M = n solute (mol) Vsolvent (L) 3 M means 3 mol/l Sample Problem 3.14 Isotonic saline is 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleansing rinse for contact lenses to a washing medium for red blood cells. How do you prepare 0.80 L of isotonic saline from 6.0 M stock solution of NaCl? M = n V
Sample Problem 3.14 Isotonic saline is 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleansing rinse for contact lenses to a washing medium for red blood cells. How do you prepare 0.80 L of isotonic saline from 6.0 M stock solution of NaCl? We ll need: n(nacl) = 0.15 mol/l! 0.80 L = 0.12 mol M = n V Sample Problem 3.14 Isotonic saline is 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleansing rinse for contact lenses to a washing medium for red blood cells. How do you prepare 0.80 L of isotonic saline from 6.0 M stock solution of NaCl? We ll need: n(nacl) = 0.15 mol/l! 0.80 L = 0.12 mol V(6M_soln) = 0.12 mol 6.0 mol/l = 0.02 L M = n V 0.78 L
Example *3.141. 50 ml of a solution of NaOH with an unknown concentration reacts with an excess of an acid solution of oxalic acid prepared by dissolving 0.588 g of oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O) in water. The excess of oxalic acid then reacts completely with 9.65 ml of 0.115 M of stock NaOH solution. What is the concentration of the unknown NaOH solution. 2NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O M = n V Example *3.141. 50 ml of a solution of NaOH with an unknown concentration reacts with an excess of an acid solution of oxalic acid prepared by dissolving 0.588 g of oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O) in water. The excess of oxalic acid then reacts completely with 9.65 ml of 0.115 M of stock NaOH solution. What is the concentration of the unknown NaOH solution. MW oad = 126.08 2NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O MW NaOH = 40.00 M = n V
Example *3.141. 50 ml of a solution of NaOH with an unknown concentration reacts with an excess of an acid solution of oxalic acid prepared by dissolving 0.588 g of oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O) in water. The excess of oxalic acid then reacts completely with 9.65 ml of 0.115 M of stock NaOH solution. What is the concentration of the unknown NaOH solution. MW oad = 126.08 2NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O MW NaOH = 40.00 n(h 2 C 2 O 4 ) = 0.588g 126.08g/mol = 4.66 mmol M = n V Example *3.141. 50 ml of a solution of NaOH with an unknown concentration reacts with an excess of an acid solution of oxalic acid prepared by dissolving 0.588 g of oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O) in water. The excess of oxalic acid then reacts completely with 9.65 ml of 0.115 M of stock NaOH solution. What is the concentration of the unknown NaOH solution. MW oad = 126.08 2NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O MW NaOH = 40.00 M = n V n(h 2 C 2 O 4 ) = 0.588g 126.08g/mol = 4.66 mmol n(naoh-known) = 9.65 ml! 0.115mol/L = 1.11 mmol
Example *3.141. 50 ml of a solution of NaOH with an unknown concentration reacts with an excess of an acid solution of oxalic acid prepared by dissolving 0.588 g of oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O) in water. The excess of oxalic acid then reacts completely with 9.65 ml of 0.115 M of stock NaOH solution. What is the concentration of the unknown NaOH solution. MW oad = 126.08 2NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O MW NaOH = 40.00 n(h 2 C 2 O 4 ) = 0.588g 126.08g/mol = 4.66 mmol n(naoh-known) = 9.65 ml! 0.115mol/L = 1.11 mmol M = n V n(oxalic-unknown) = 4.66-1.11/2 = 4.11 mmol Example *3.141. 50 ml of a solution of NaOH with an unknown concentration reacts with an excess of an acid solution of oxalic acid prepared by dissolving 0.588 g of oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O) in water. The excess of oxalic acid then reacts completely with 9.65 ml of 0.115 M of stock NaOH solution. What is the concentration of the unknown NaOH solution. MW oad = 126.08 2NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O MW NaOH = 40.00 n(naoh-unknown) = 4.11mmol!2 = 8.22 mmol M = n V M = 8.22 mmol 50 ml = 0.164 M Look at equation
Chapter 3 Overview Stoichiometry The mole: proportional to the number of molecules a convenient measure of amount n(mol) = m(g) / M(g/mol) M = MW Chemical equations: balanced: same number of atoms on both sides show stoichiometric relationships between reactants and products Solutions: concentration M (mol/l) = n(mol) / V(L) Chapter 3 Overview Practice Problems (3.44) Cortisol (M = 362.47 g/mol), one of the major steroid hormones, is a key factor in the biosynthesis of protein. Cortisol is 69.6% carbon, 8.34% H, 22.1% O by mass. What is its molecular formula?
Chapter 3 Overview Practice Problems (3.44) Cortisol (M = 362.47 g/mol), one of the major steroid hormones, is a key factor in the biosynthesis of protein. Cortisol is 69.6% carbon, 8.34% H, 22.1% O by mass. What is its molecular formula? (1) get molar ratio of C H O " get empirical formula (2) look at molecular weight, empirical formula " get molecular formula Chapter 3 Overview Practice Problems (3.51) Balance the following equations: (a) Cu + S 8 # Cu 2 S
Chapter 3 Overview Practice Problems (3.51) Balance the following equations: (a) Cu + S 8 # Cu 2 S 16Cu + S 8 # 8Cu 2 S Chapter 3 Overview Practice Problems (3.51) Balance the following equations: (a) Cu + S 8 # Cu 2 S 16Cu + S 8 # 8Cu 2 S (b) P 4 O 10 + H 2 O # H 3 PO 4
Chapter 3 Overview Practice Problems (3.51) Balance the following equations: (a) Cu + S 8 # Cu 2 S 16Cu + S 8 # 8Cu 2 S (b) P 4 O 10 + H 2 O # H 3 PO 4 (3.52) Balance the following equations: (c) CaSiO 3 + HF # SiF 4 + CaF 2 + H 2 O Chapter 3 Overview Practice Problems (3.56) Convert the following into balanced equations: (a) When lead (II) nitrate solution is added to potassium iodide solution, solid lead (II) iodide forms and potassium nitrate solution remains.
Chapter 3 Overview Practice Problems (3.56) Convert the following into balanced equations: (a) When lead (II) nitrate solution is added to potassium iodide solution, solid lead (II) iodide forms and potassium nitrate solution remains. Pb(NO 3 ) 2 + 2KI # PbI 2 + 2KNO 3 Chapter 3 Overview Practice Problems ( 3.61) Chlorine gas can be made in the laboratory by the reaction of hydrochloric acid and manganese (IV) dioxide: 4HCl + MnO 2 # MnCl 2 + Cl 2 + 2H 2 O When 1.82 mol of HCl reacts with excess MnO 2, (a) how many mols of Cl 2 form? (b) How many grams of Cl 2 form?
Chapter 3 Overview Practice Problems ( 3.61) Chlorine gas can be made in the laboratory by the reaction of hydrochloric acid and manganese (IV) dioxide: 4HCl + MnO 2 # MnCl 2 + Cl 2 + 2H 2 O When 1.82 mol of HCl reacts with excess MnO 2, (a) how many mols of Cl 2 form? (b) How many grams of Cl 2 form? n (mol) = m (g) M (g/mol) Chapter 3 Overview Practice Problems ( 3.74) Calculate the maximum numbers of moles and grams of H 2 S that can form when 158 g of aluminum sulfide reacts with 131 g of water: Al 2 S 3 + H 2 O # Al(OH) 3 + H 2 S [unbalanced] What mass of the excess reactant remains?
Chapter 3 Overview Practice Problems: Solutions ( 3.93) Calculate each of the following quantities: (a) Volume in liters of 2.26 M potassium hydroxide that contains 8.42 g of solute. (c) molarity of 275 ml solution containing 135 mmol of glucose. Chapter 3 Overview Practice Problems: Solutions ( 3.103) How many moles of which reactant are in excess when 350.0 ml of 0.210 M sulfuric acid reacts with 0.500 L of 0.196 M sodium hydroxide to form water and aqueous sodium sulfate.
Chapter 3 Overview Practice Problems:Comprehensive ( 3.133) To 1.20 L of 0.325 M HCl, you add 3.37 L of a second HCl solution of unknown concentration. The resulting solution is 0.893 M HCl. Assuming the volumes are additive, calculate the molarity of the second HCl solution. Chapter 3 Overview Practice Problems:Comprehensive ( *3.141) In the chemical analysis of unknown, chemsits often add an excess of a reactant, determine the amount of that reactant remaining after the reaction with the unknown, and use those amounts to calculate the amount of the unknown. For analysis of an unknown NaOH solution, you add 50.0 ml of the solution to 0.150 L of an acid solution prepared by dissolving 0.588 g of solid oxalic acid dihydrate (H 2 C 2 O 4 2H 2 O, MW = 126.08) in water: 2NaOH + H 2 C 2 O 4 # Na 2 C 2 O 4 + 2H 2 O The unreacted NaOH then reacts completely with 9.65 ml of 0.116 M HCl: NaOH + HCl # NaCl + H 2 O What is the molarity of the original NaOH solution?