Solving Stoichiometry Problems for Reactions in Solution

Section 4.7 Stoichiometry of Precipitation Reactions Solving Stoichiometry Problems for Reactions in Solution 1. Determine what reaction if any occurs. If a reaction occurs write a balanced molecular equation. 2. Write the balanced net ionic equation for the reaction. 3. Calculate the moles of reactants. 4. Determine which reactant is limiting. (if given 2 reactants) 5. Convert to moles of product & then desired unit Copyright Cengage Learning. All rights reserved 1

Section 4.7 Stoichiometry of Precipitation Reactions Stoichiometry of Precipitation Reactions: Example 1: Calculate the mass of barium sulfate formed when 250.0 ml of 0.0500 M BaCl 2 and 150.0 ml of 0.100 M Na 2 SO 4 are mixed. Step 1: The Molecular equation: BaCl 2 + Na 2 SO 4 BaSO 4 + 2NaCl Step 2: The net Ionic equation Ba 2+ + SO 4 2- BaSO 4 Copyright Cengage Learning. All rights reserved 2

Section 4.7 Stoichiometry of Precipitation Reactions Step 3: Calculate the moles of reactants 1L 0.0500mol mol Ba 250.0 ml 0.0125 mol Ba 1000mL L 2 2 1L 0.100mol mol SO 150.0 ml 0.0150 mol SO 1000mL L Step 4: Determine the limiting reagent 0.0150 2 2 4 4 2 mol SO 4 1 0.0150 0.0125 2 mol Ba 0.0125 Step 5: Determine the moles of product and then convert to other units as needed 1 mol BaSO 233.4 g BaSO 0.0125 mol Ba 2.92 g BaSO 2 4 4 2 1 mol Ba 1 mol BaSO4 Copyright Cengage Learning. All rights reserved 3 1 Ba 2+ is limiting 4

Section 4.7 Stoichiometry of Precipitation Reactions Example 2: 10.0 ml of a 0.30 M sodium phosphate solution reacts with 20.0 ml of a 0.20 M lead(ii) nitrate solution (assume no volume change) What mass of precipitate will form? Let s Think About It Where are we going? Find the mass of solid Pb 3 (PO 4 ) 2 formed. How do we get there? Write the balanced molecular & net ionic equation What are the moles of reactants present in the solution? Which reactant is limiting? Convert to moles of product & then to the desired unit What mass of Pb 3 (PO 4 ) 2 will be formed? Copyright Cengage Learning. All rights reserved 4

Section 4.7 Stoichiometry of Precipitation Reactions Example 2: 10.0 ml of a 0.30 M sodium phosphate solution reacts with 20.0 ml of a 0.20 M lead(ii) nitrate solution (assume no volume change) What mass of precipitate will form? Step 1: The Molecular equation: (what ppt forms) 2Na 3 PO 4 (aq) + 3Pb(NO 3 ) 2 (aq) 6NaNO 3 (aq) + Pb 3 (PO 4 ) 2 (s) Step 2: The net Ionic equation 3Pb 2+ + 2PO 4 3- Pb 3 (PO 4 ) 2 Copyright Cengage Learning. All rights reserved 5

Section 4.7 Stoichiometry of Precipitation Reactions Step 3: Calculate the moles of reactants 1L 0.30mol 10.0 mlx x.0030molna PO 1000mL 1L Step 4: Determine the limiting reagent 3 4 1L 0.20mol 20.0 mlx x.0040 molpb(n O ) 1000mL 1L.0030 molna 3 PO 4 3 2 2 0.0015.0040 molpb(n O ).0013 3 Step 5: Determine moles of product & convert to the desired unit 1 molpb3( PO4) 2 811.5 gpb3( PO4) 2.0040x x 1.1g Pb ( PO ) 3 molpb(n O3) 2 1 molpb3( PO4) 2 3 2 Pb 2+ is limiting. 3 4 2 Copyright Cengage Learning. All rights reserved 6

Section 4.7 Stoichiometry of Precipitation Reactions Example 3: 10.0 ml of a 0.30 M sodium phosphate solution reacts with 20.0 ml of a 0.20 M lead(ii) nitrate solution (assume no volume change). What is the concentration of nitrate ions left in solution after the reaction is complete? Let s Think About It Where are we going? To find the concentration of nitrate ions left in solution after the reaction is complete. How do we get there? What are the moles of nitrate ions present in the combined solution? What is the total volume of the combined solution? Copyright Cengage Learning. All rights reserved 7

Section 4.7 Stoichiometry of Precipitation Reactions Example 3: Nitrate ions are spectator ions & don t participate in the reaction. We know from the earlier problem that Pb +2 was limiting therefore:.0040m.0040 NO x2.0080mol NO 3 3 The total volume of the solution is 30.0mL.0080mol N.0300L O 3.27M N O 3 Copyright Cengage Learning. All rights reserved 8

Section 4.7 Stoichiometry of Precipitation Reactions Example 4: 10.0 ml of a 0.30 M sodium phosphate solution reacts with 20.0 ml of a 0.20 M lead(ii) nitrate solution (assume no volume change). What is the concentration of phosphate ions left in solution after the reaction is complete? Let s Think About It Where are we going? Find the concentration of phosphate ions left after the reaction is complete. How do we get there? What are the moles of phosphate ions present in the solution at the start of the reaction? How many moles of phosphate ions were used up in the reaction to make the solid Pb 3 (PO 4 ) 2? How many moles of phosphate ions are left over after the reaction is complete? What is the total volume of the combined solution? Copyright Cengage Learning. All rights reserved 9

Section 4.7 Stoichiometry of Precipitation Reactions Example 4: Take moles of limiting and turn to excess; this determines the amount of phosphate used during the reaction. 2molPO 4 3.0040 Pb+2 x.0027molpo 4 3 3molPb 2 There were 0.0030 mol of Na 3 PO 4 at the start, which means there are 0.0030 mol PO 4-3 0.0030 mol 0.0027 mol= 0.00033 mol of PO 4-3 left Copyright Cengage Learning. All rights reserved 10

Section 4.7 Stoichiometry of Precipitation Reactions Example 5: Calculate the mass of Ag 2 S produced when 125ml of 0.200MAgNO 3 is added to an excess of Na 2 S solution. 2AgNO 3 + Na 2 S Ag 2 S +2NaNO 3 1 L.200molAgNO 3 1molAg 2S 247.9gAg 2S 125mLAgNO 3x x x x 3.10gAg 2S 1000mL 1LAgNO 3 2molAgNO 3 1molAg 2S Copyright Cengage Learning. All rights reserved 11

Section 4.8 Acid Base Reactions Acid Base Reactions (Brønsted Lowry) Acid proton donor Base proton acceptor For a strong acid and base reaction: H + (aq) + OH (aq) H 2 O(l) Copyright Cengage Learning. All rights reserved 13

Section 4.8 Acid Base Reactions Performing Calculations for Acid Base Reactions 1. List the species present in the combined solution before any reaction occurs, and decide what reaction will occur. 2. Write the balanced net ionic equation for this reaction. 3. Calculate moles of reactants. 4. Determine the limiting reactant, where appropriate. 5. Calculate the moles of the required reactant or product. 6. Convert to grams or volume (of solution), as required. Copyright Cengage Learning. All rights reserved 15

Section 4.8 Acid Base Reactions Acid Base Titrations Titration delivery of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte). Equivalence point enough titrant added to react exactly with the analyte. Endpoint the indicator changes color so you can tell the equivalence point has been reached. Copyright Cengage Learning. All rights reserved 16

Section 4.8 Acid Base Reactions Concept Check For the titration of sulfuric acid (H 2 SO 4 ) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to react with 1.00 L of 0.500 M sulfuric acid to reach the endpoint? 1.00 mol NaOH Copyright Cengage Learning. All rights reserved 17

Section 4.8 Acid Base Reactions Let s Think About It Where are we going? To find the moles of NaOH required for the reaction. How do we get there? What are the ions present in the combined solution? What is the reaction? What is the balanced net ionic equation for the reaction? What are the moles of H + present in the solution? How much OH is required to react with all of the H + present? Copyright Cengage Learning. All rights reserved 18

Section 4.9 Oxidation Reduction Reactions Rules for Assigning Oxidation States 1. Oxidation state of an atom in an element = 0 2. Oxidation state of monatomic ion = charge of the ion 3. Oxygen = 2 in covalent compounds (except in peroxides where it = 1) 4. Hydrogen = +1 in covalent compounds 5. Fluorine = 1 in compounds 6. Sum of oxidation states = 0 in compounds 7. Sum of oxidation states = charge of the ion in ions Copyright Cengage Learning. All rights reserved 21

Section 4.9 Oxidation Reduction Reactions Exercise Find the oxidation states for each of the elements in each of the following compounds: K 2 Cr 2 O 7 CO 3 2- MnO 2 PCl 5 SF 4 K = +1; Cr = +6; O = 2 C = +4; O = 2 Mn = +4; O = 2 P = +5; Cl = 1 S = +4; F = 1 Copyright Cengage Learning. All rights reserved 22

Section 4.9 Oxidation Reduction Reactions Redox Characteristics Transfer of electrons Transfer may occur to form ions Oxidation increase in oxidation state (loss of electrons); reducing agent Reduction decrease in oxidation state (gain of electrons); oxidizing agent Copyright Cengage Learning. All rights reserved 23

Section 4.9 Oxidation Reduction Reactions Concept Check Which of the following are oxidation-reduction reactions? Identify the oxidizing agent and the reducing agent. a)zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) b)cr 2 O 7 2- (aq) + 2OH - (aq) 2CrO 4 2- (aq) + H 2 O(l) c)2cucl(aq) CuCl 2 (aq) + Cu(s) Copyright Cengage Learning. All rights reserved 24

Section 4.10 Balancing Oxidation Reduction Equations Balancing Oxidation Reduction Reactions by Oxidation States 1. Write the unbalanced equation. 2. Determine the oxidation states of all atoms in the reactants and products. 3. Show electrons gained and lost using tie lines. 4. Use coefficients to equalize the electrons gained and lost. 5. Balance the rest of the equation by inspection. 6. Add appropriate states. Copyright Cengage Learning. All rights reserved 25

Section 4.10 Balancing Oxidation Reduction Equations Balance the reaction between solid zinc and aqueous hydrochloric acid to produce aqueous zinc(ii) chloride and hydrogen gas. Copyright Cengage Learning. All rights reserved 26

Section 4.10 Balancing Oxidation Reduction Equations 2. What are the oxidation states for each atom? Zn(s) + HCl(aq) Zn 2+ (aq) + Cl (aq) + H 2 (g) 0 +1 1 +2 1 0 Copyright Cengage Learning. All rights reserved 28

Section 4.10 Balancing Oxidation Reduction Equations 3. How are electrons gained and lost? 1 e gained (each atom) Zn(s) + HCl(aq) Zn 2+ (aq) + Cl (aq) + H 2 (g) 0 +1 1 +2 1 0 2 e lost The oxidation state of chlorine remains unchanged. Copyright Cengage Learning. All rights reserved 29

Section 4.10 Balancing Oxidation Reduction Equations 4. What coefficients are needed to equalize the electrons gained and lost? 1 e gained (each atom) 2 Zn(s) + HCl(aq) Zn 2+ (aq) + Cl (aq) + H 2 (g) 0 +1 1 +2 1 0 2 e lost Zn(s) + 2HCl(aq) Zn 2+ (aq) + Cl (aq) + H 2 (g) Copyright Cengage Learning. All rights reserved 30

Section 4.10 Balancing Oxidation Reduction Equations 5. What coefficients are needed to balance the remaining elements? Zn(s) + 2HCl(aq) Zn 2+ (aq) + 2Cl (aq) + H 2 (g) Copyright Cengage Learning. All rights reserved 31