) DON T FORGET WHAT THIS REPRESENTS

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1 6/.A Stoichiometry: Molarity I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS a zero ( 0 ) is better for you than copying Instructions: ON A SEPARATE SHEET OF PAPER, given the following data, perform the following molarity calculations. Remember, Sig Figs, NW = NC, Boxed Answers and N 3 (this includes labeling and in your set-up!!!) 1. Sea water contains roughly 28.0 g of NaCl per liter. What is the molarity of sodium chloride in sea water? Plan: g / L NaCl mol NaCl MMNaCl = g/mol NaCl MM = Read it from PT 10. Determine the molarity of these solutions: a moles of Li2SO3 dissolved to make 2.04 liters of solution. Plan: mol / L Li2SO g NaCl 1 mol NaCl = M NaCl3 1 L g NaCl 2. What is the molarity of g of H2SO4 dissolved in 1.00 L of solution? Plan: g / L H2SO4 mol H2SO4 MM H2SO4 = g/mol H2SO4 MM = Read it from PT g H2SO4 1 mol H2SO4 = M H2SO4 1 L g H2SO4 3. What is the molarity of 5.30 g of Na2CO3 dissolved in ml solution? Plan: g / L Na2CO3 mol Na2CO3 MM Na2CO3 = g/mol Na2CO3 MM = Read it from PT 5.30 g Na2CO3 1 mol Na2CO3 = M Na2CO L g Na2CO3 4. What is the molarity of 5.00 g of NaOH in ml of solution? Plan: g / L NaOH mol NaOH MM NaOH = g/mol NaOH MM = Read it from PT 5.00 g NaOH 1 mol NaOH = M NaOH L g NaOH 5. How many moles of Na2CO3 are there in 10.0 L of 2.0 M solution? Plan: L 2.0 M Na2CO3 mol Na2CO L Na2CO3 2.0 mol Na2CO3 = 20.0 mol Na2CO3 1 L Na2CO3 6. What mass of H2SO4 would be needed to make 75 ml of 2.00 M solution? Plan: g / L H2SO4 mol H2SO4 g H2SO4 MM H2SO4 = g/mol H2SO L H2SO mol H2SO g H2SO4 1 L H2SO4 1 mol H2SO4 = 15 g H2SO4 7. What volume (in ml) of 18.0 M H2SO4 is needed to contain 2.45 g H2SO4? Plan: g H2SO4 mol H2SO4 L 18.0 M H2SO4 ml 18.0 M H2SO4 MM H2SO4 = g/mol H2SO g H2SO4 1 mol H2SO4 1 L H2SO ml g H2SO mol H2SO4 1 L = 1.39 ml 18.0 M H2SO4 8. What volume (in ml) of 12.0 M HCl is needed to contain 3.00 moles of HCl? Plan: mol HCl L 12.0 M HCl ml 12.0 M HCl 3.00 mol HCl 1 L HCl 1000 ml 12.0 mol HCl 1 L = 250. ml 12.0 M HCl 9. What mass of KCl is there in 2.50 liters of 0.50 M KCl solution? Plan: L 0.50 M KCl mol KCl g KCl MM KCl = g/mol KCl 2.50 L KCl 0.50 mol KCl g KCl 1 L KCl 1 mol KCl = 93.2 g KCl 4.67 mol Li2SO3 = 2.29 M Li2SO L b moles of Al2O3 to make liters of solution. Plan: mol / L Al2O mol Al2O3 = M Al2O L c grams of Na2CO3 to make liters of solution. Plan: g / L Na2CO3 mol Na2CO3 MM Na2CO3 = g/mol Na2CO g Na2CO3 1 mol Na2CO3 = M Na2CO L g Na2CO3 d grams of (NH4)2CO3 to make 250 ml of solution. Plan: g / L (NH4)2CO3 mol (NH4)2CO3 MM (NH4)2CO3= g/mol (NH4)2CO g (NH4)2CO3 1 mol (NH4)2CO3 = M (NH4)2CO L g (NH4)2CO3 11. Determine the number of moles of solute to prepare these solutions: a liters of a 2.00 M Cu(NO3)2 solution. Plan: L 2.00 M Na2CO3 mol Na2CO L Cu(NO3) mol Cu(NO3)2 = 4.70 mol Cu(NO3)2 1 L Cu(NO3)2 b ml of a molar Pb(NO3)2 solution. Plan: L M Pb(NO3)2 mol Pb(NO3) L Pb(NO3) mol Pb(NO3)2 = mol Pb(NO3)2 1 L Pb(NO3)2 c L of a M MgCO3 solution. Plan: L M MgCO3 mol MgCO L MgCO mol MgCO3 = 1.50 mol MgCO3 1 L MgCO3 d L of a 3.76-molar Na2O solution. Plan: L 3.76 M Na2O mol Na2O 6.20 L Na2O 3.76 mol Na2O = 23.3 mol Na2O 1 L Na2O

2 12. Determine the grams of solute to prepare these solutions: a liters of a M Cu(NO3)2 solution. Plan: L M Cu(NO3)2 mol Cu(NO3)2 g Cu(NO3)2 MM Cu(NO3)2 = g/mol Cu(NO3) L Cu(NO3) mol Cu(NO3) g Cu(NO3)2 1 L Cu(NO3)2 1 mol Cu(NO3)2 = g Cu(NO3)2 b milliliters of a 5.90-molar Pb(NO3)2 solution. Plan: L 5.90 M Pb(NO3)2 mol Pb(NO3)2 g Pb(NO3)2 MM Pb(NO3)2 = g/mol Pb(NO3) L Pb(NO3) mol Pb(NO3) g Pb(NO3)2 1 L Pb(NO3)2 1 mol Pb(NO3)2 = 31.3 g Pb(NO3)2 c. 508 ml of a 2.75-molar NaF solution. Plan: L 2.75 M NaF mol NaF g NaF MM NaF = g/mol NaF L NaF 2.75 mol NaF g NaF 1 L NaF 1 mol NaF = 58.7g NaF d L of a 3.76-molar Na2O solution. Plan: L 3.76 M Na2O mol Na2O g Na2O MM Na2O = g/mol Na2O 6.20 L Na2O 3.76 mol Na2O g Na2O 1 L Na2O 1 mol Na2O = 1440 g Na2O 13. Determine the final volume of these solutions: a moles of Li2SO3 dissolved to make a 3.89 M solution. Plan: mol Li2SO3 L 3.98 M Li2SO mol Li2SO3 1 L Li2SO3 = 1.20 L 3.89 M Li2SO mol Li2SO3 b moles of Al2O3 to make a M solution. Plan: mol Al2O3 L M Al2O mol Al2O3 1 L Al2O3 = L M Li2SO mol Al2O3 c grams of Na2CO3 to make a M solution. Plan: g Na2CO3 mol Na2CO3 L M Na2CO3 MM Na2CO3 = g/mol Na2CO g Na2CO3 1 mol Na2CO3 1 L Na2CO g Na2CO mol Na2CO3 = L of M Na2CO3 d grams of (NH4)2CO3 to make a molar solution. Plan: g (NH4)2CO3 mol (NH4)2CO3 L M (NH4)2CO3 MM (NH4)2CO3= g/mol (NH4)2CO g (NH4) 2CO3 1 mol (NH4) 2CO3 1 L (NH4)2CO g (NH4) 2CO mol (NH4) 2CO3 = L of M (NH4)2CO3 Molarity Resource Best Option Alternative Stoichiometry Resources Bozeman Review Bozeman Walk-Through 6.B Stoichiometry: Mole Map Basics I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS a zero ( 0 ) is better for you than copying Instructions: Given the following chemical equations, balance them. Then ON A SEPARATE SHEET OF PAPER, use dimensional analysis to perform the following stoichiometric calculations. You MUST use dimensional analysis to preform each conversion. Remember, Sig Figs, NW = NC, Boxed Answers and N 3 (this includes labeling and in your set-up!!!) 1. 2 NaCl(aq) + F2(g) 2 NaF(aq) + Cl2(g) a mol F2 L of F2 Plan: mol F2 (g) L F2 (g) Remember, this is a gas. YOU HAVE TO KNOW that 1 mol of gas = 22.4 STP 0.32 mol F L F2 (g) = 7.2 L F2 (g) 1 mol F2 b L of 2.0 M NaCl g NaCl Plan: L 2.0 M NaCl mol NaCl g NaCl MM NaCl = 1 x g/mol Na + 1 x g/mol Cl = g/mol NaCl 5.61 L NaCl 2.0 mol NaCl(aq) g NaCl = 656 g NaCl 1 L NaCl 1 mol NaCl c g Cl2 L Cl2 Plan: g Cl2 (g) mol Cl2 (g) L Cl2 (g) MM Cl2 = 2 x g/mol Cl = Cl2 Remember, this is a gas. YOU HAVE TO KNOW that 1 mol of gas = 22.4 STP g Cl2 1 mol Cl L Cl2 (g) = L Cl2 (g) g Cl2 1 mol Cl2

3 2. Pb(OH)2(aq)+ 2 HCl(aq) 2 H2O(l) + PbCl2 (s) a mol Pb(OH)2 mol of H2O Plan: mol Pb(OH)2 mol H2O 2.42 mol Pb(OH)2 2 mol H2O = 4.84 mol H2O b. 120 mol HCl mol PbCl2 Plan: mol HCl mol PbCl2 1 mol Pb(OH)2 120 mol HCl 1 mol PbCl2 = 60. mol PbCl2 2 mol HCl c x 10 7 mol PbCl2 mol Pb(OH)2 Plan: mol PbCl2 mol Pb(OH) x 10 7 mol PbCl2 1 mol Pb(OH)2 = 2.02 x 10 7 mol Pb(OH)22. 1 mol PbCl AlBr3 (aq) + 3 K2SO4 (aq) 6 KBr (aq) + Al2(SO4)3(s) a g K2SO4 g of KBr Plan: g K2SO4 mol K2SO4 mol KBr g KBr MM K2SO4 = 2 x g/mol K + 1 x g/mol S + 4 x g/mol S = g/mol K2SO4 MM KBr = 1 x g/mol K + 1 x g/mol Br = g/mol KBr g K2SO4 1 mol K2SO4 6 mol KBr g KBr = g KBr g K2SO4 3 mol K2SO4 1 mol KBr b g AlBr3 g Al2(SO4)3 Plan: g AlBr3 mol AlBr3 mol Al2(SO4)3 g Al2(SO4)3 MM AlBr2 = 1 x g/mol Al + 3 x g/mol Br = g/mol AlBr3 MM Al2(SO4)3 = 2 x g/mol Al + 3 x g/mol S + 12 x g/mol O = g/mol Al2(SO4) g AlBr3 1 mol AlBr3 1 mol Al2(SO4) g Al2(SO4)3 = g Al2(SO4) g AlBr3 2 mol AlBr3 1 mol Al2(SO4)3 c g KBr g K2SO4 Plan: g KBr mol KBr mol K2SO4 g K2SO4 MM K2SO4 = 2 x g/mol K + 1 x g/mol S + 4 x g/mol S = g/mol K2SO4 MM KBr = 1 x g/mol K + 1 x g/mol Br = g/mol KBr g KBr 1 mol KBr 3 mol K2SO g K2SO4 = 5.51x 10 4 g K2SO g KBr 6 mol KBr 1 mol K2SO C7H10 (s) + 19 O2 (g) 14 CO2 (g) + 10 H2O (g) a L O2 L CO2 Plan: L O2 (g) L CO2 Remember, 1 L gas = 1 L STP L O2 14 L CO2 = L CO2 19 L O2 b L H2O mol C7H10 Plan: L H2O (g) mol H2O (g) mol C7H10 Remember, this is a gas. YOU HAVE TO KNOW that 1 mol of gas = 22.4 STP L H2O 1 mol H2O 2 mol C7H10 = mol C7H L of H2O(g) 10 mol H2O c L CO2 g C7H10 Plan: L H2O (g) mol H2O (g) mol C7H10 MM C7H10 = 7 x g/mol C + 10 x 1.01 g/mol H = g/mol C7H10 Remember, this is a gas. YOU HAVE TO KNOW that 1 mol of gas = 22.4 STP 5.61 L CO2 1 mol CO2 2 mol C7H g C7H10 = 3.37 g C7H L of CO2 (g) 14 mol CO2 1 mol C7H10

4 5. FeCl3 (aq) + 3 NaOH (aq) Fe(OH)3(s) + 3 NaCl(aq) a g NaCl g of Fe(OH)3 Plan: g NaCl mol NaCl mol Fe(OH)3 g Fe(OH)3 MM NaCl = 1 x g/mol Na + 1 x g/mol Cl = g/mol NaCl MM Fe(OH)3 = 1 x g/mol Fe + 3 x g/mol O + 3 x 1.01 g/mol H = g/mol Fe(OH) g NaCl 1 mol NaCl 1 mol Fe(OH) g Fe(OH)3 = g Fe(OH)3 b mol Fe(OH)3 L of 1.24 M NaCl Plan: mol Fe(OH)3 mol NaCl(aq) L 1.24 M NaCl g NaCl 3 mol NaCl 1 mol Fe(OH) mol Fe(OH)3 3 mol NaCl (aq) 1 L of NaCl (aq) = 23.3 L of 1.24 M NaCl 1 mol Fe(OH) mol NaCl c L of 4.5 M NaOH L of 6.0 M FeCl3 Plan: L 4.5 M NaOH mol NaOH mol FeCl3 L 6.0 M FeCl L NaOH 4.5 mol NaOH 1 mol FeCl3 1 L FeCl3 = L of 6.0 M FeCl Ag2O(s) 4 Ag(s) + O2(g) a L O2 g Ag 1L NaOH 3 mol NaOH 6.0 mol FeCl3 Plan: L O2 (g) mol O2 (g) mol Ag g Ag MM Ag = g/mol Ag Remember, this is a gas. YOU HAVE TO KNOW that 1 mol of gas = 22.4 STP L O2 1 mol O2 4 mol Ag g Ag = 1023 g Fe(OH) L O2 (g) 1 mol O2 1 mol Ag b g Ag2O L O2 Plan: g Ag2O mol Ag2O mol O2 (g) L O2 (g) MM Ag2O = 2 x g/mol Ag + 1 x g/mol O = g/mol Ag2O Remember, this is a gas. YOU HAVE TO KNOW that 1 mol of gas = 22.4 STP g Ag2O 1 mol Ag2O 1 mol O L O2 = L O2 (g) c g Ag FU Ag2O g Ag2O 2 mol Ag2O 1 mol O2 Plan: g Ag mol Ag mol Ag2O FU Ag2O MM Ag = g/mol Ag Remember, 1 mol = x particles. YOU HAVE TO KNOW that 1 FU is equivalent to 1 particle 2300 g Ag 1 mol Ag 2 mol Ag2O 6.022x10 23 FU Ag2O = 6.4 x Ag2O g Ag 4 mol Ag 1 mol Ag2O 7. 2 K(s) + MgBr2 (aq) 2 KBr (aq) + Mg(s) a x atoms K FU MgBr2 Plan: atoms K FU MgBr2 Yes, you could go to moles, but you do not need to in this case x atoms K 1 FU MgBr2 = 7.10 x FU MgBr2 b L of 0.53 M MgBr2 Atoms Mg 2 atoms K Plan: L 0.53 M MgBr2 mol MgBr2 mol Mg FU Mg Remember, 1 mol = x particles. YOU HAVE TO KNOW that 1 atom is equivalent to 1 particle 5.24 L MgBr mol MgBr2 1 mol Mg 6.022x10 23 atoms Mg = 1.67 x atoms Mg c. 8.8 x FU KBr g K 1 L MgBr2 1 mol MgBr2 1 mol Mg Plan: FU KBr mol KBr mol K g K MM K = g/mol K Remember, 1 mol = x particles. YOU HAVE TO KNOW that 1 FU is equivalent to 1 particle 8.8 x FU KBr 1 mol KBr 2 mol K g K = 5700 g K 6.022x10 23 FU KBr 2 mol KBr 1 mol K 8. 2 HCl(aq) + CaCO3(s) CaCl2(aq) + H2CO3(aq) a L of 0.53 M HCl g CaCl2 Plan: L 0.53 M HCl mol HCl mol CaCl2 g CaCl2 MM CaCl2 = 1 x g/mol Ca + 2 x g/mol Cl = g/mol CaCl L HCl 0.53 mol HCl 1 mol CaCl g CaCl2 = 13 g CaCl2 1 L HCl 2 mol HCl 1 mol CaCl2 b x FU CaCO3 L of M CaCl2 Plan: FU CaCO3 mol CaCO3 mol CaCl2 L M CaCl2 Remember, 1 mol = x particles. YOU HAVE TO KNOW that 1 atom is equivalent to 1 particle 5.61 x FU CaCO3 1 mol CaCO3 1 mol CaCl2 1 L CaCl2 = 17.9 L of M CaCl2 c g CaCl2 FU H2CO x10 23 FU CaCO3 1 mol CaCO mol CaCl2 Plan: g CaCl2 mol CaCl2 mol H2CO3 FU H2CO3 MM CaCl2 = 1 x g/mol Ca + 2 x g/mol Cl = g/mol CaCl2 Remember, 1 mol = x particles. YOU HAVE TO KNOW that 1 FU is equivalent to 1 particle 4.54 g CaCl2 1 mol CaCl2 1 mol H2CO x10 23 FU H2CO3 = 2.46 x FU H2CO g CaCl2 1 mol CaCl2 1 mol H2CO3

5 9. 2 NaBr (aq) + CaF2 (aq) 2 NaF(aq) + CaBr2(s) a g CaF2 g NaF Plan: g CaF2 mol CaF2 mol NaF g NaF MM NaF = 1 x g/mol Na + 1 x g/mol F = g/mol NaF MM CaF2 = 1 x g/mol Ca + 2 x g/mol F = g/mol CaF g CaF2 1 mol CaF2 2 mol NaF g NaF = g NaF g CaF2 1 mol CaF2 1 mol NaF b x FU NaBr FU CaBr2 Plan: FU NaBr FU CaBr2 Yes, you could go to moles, but you do not need to in this case x FU NaBr 1 FU CaBr2 = 2.56 x FU CaBr2 2 FU NaBr c mol NaF FU CaF2 Plan: mol NaF mol CaF2 FU CaF2 Remember, 1 mol = x particles. YOU HAVE TO KNOW that 1 FU is equivalent to 1 particle mol NaF 1 mol CaF x10 23 FU CaF2 = 2.80 x FU CaF2 2 mol NaF 1 mol CaF2 10. H2SO4(aq) + 2 NaNO2(aq) 2 HNO2(aq) + Na2SO4(aq) a L of M H2SO4 g HNO2 Plan: L M H2SO4 mol H2SO4 mol HNO2 g HNO2 MM HNO2 = 1 x 1.01 g/mol H + 1 x g/mol N + 2 x g/mol O = g/mol HNO L H2SO mol H2SO4 2 mol HNO g HNO2 = 25.2 g HNO2 b. 930 mol NaNO2 g H2SO4 1 L H2SO4 1 mol H2SO4 1 mol HNO2 Plan: mol NaNO2 mol H2SO4 g H2SO4 MM H2SO4 = 2 x 1.01 g/mol H + 1 x g/mol S + 4 x g/mol O = g/mol H2SO4 930 mol NaNO2 1 mol H2SO g H2SO4 = g H2SO4 c x FU HNO2 L of 0.31 M Na2SO4 2 mol NaNO2 1 mol H2SO4 Plan: FU HNO2 mol HNO2 mol Na2SO4 L 0.31 M Na2SO4 Remember, 1 mol = x particles. YOU HAVE TO KNOW that 1 FU is equivalent to 1 particle 9.61 x FU HNO2 1 mol HNO2 1 mol Na2SO4 1 L Na2SO4 = 2.57x10 8 L of 0.31 M Na2SO x10 23 FU HNO2 2 mol HNO mol Na2SO4 6.C Solutions: The Basics I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS a zero ( 0 ) is better for you than copying Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE. 1. A solution is defined as homogeneous mixture in a single phase which means that a solution DOES NOT just meant aqueous (liquid) 2. The two components of a solution are the solute which is what is being dissolved and the solvent which is what is doing the dissolving 3. The most common solvent is water H2O / dihydrogen monoxide making it known as the Universal Solvent. 4. What simple phrase explain how the forces of attraction between solute and solvent determines if a solution will form or the solute will remain insoluble: Like Dissolves Like meaning, polar substances (having charges) are dissolved by polar substances (as the opposite charges attract), nonpolar (having no charges) dissolve in nonpolar (as the charges of polar molecules do not attract to them and they separate) 5. What factors can be manipulated to increase the rate at which a solution will form? Increase temperature of solute/solvent Stirring/mixing increases particle interaction Increase surface are of solute/solvent 6. A compound is an electrolyte if it is ionic (a cation bonded to an anion ) and it can dissociate (NOT JUST DISSOVLE) in water. 7. What is the difference between an electrolyte and a nonelectrolyte? Electrolyte = ionic compounds that dissociate and as a result are able to conduct electricity Nonelectrolyte = molecular compounds that dissolve in water, but as they are not ionic and cannot dissociate, they are unable to conduct electricity 8. When a solute is added to water, the freezing point gets depressed (lowers) and the boiling point gets elevated (raises). Explain why this occurs: Presence of a solute decreases the IMF interactions of a substance. As water possess H-bonding, the strongest IMF the resulting new IMF interaction is weaker than that of pure water. This lowers the vapor pressure of the any temperature, making it necessary to heat a solution to a higher temp to boil and a lower temperature to freeze.

6 9. Electrolytes make a greater difference in freezing and boiling temperature than nonelectrolytes. Explain why this occurs: Electrolytes dissociate in solution, meaning more particles are added to a solution than individual molecules/fu indicate (i.e. NaCl donates 2 particles, Na + and Cl - not just one FU of NaCl). More particles = more IMF interactions = greater effect Instructions: Given the following reactants, write a complete and balanced chemical equation for the double replacment reaction using your references. Determine the ionic composition of each formula, do your double replacment (flip-flop cations) and apply solubility rules. RXN only occurs if one or more substance is no longer aqueous. Otherwise NO RXN occurs Ca(OH)2(aq) H3PO4(aq) 6 H2O (l) + Ca3(PO4)2 (s) Ions: Ca 2+ OH H + PO4 3 Water = Pure = l aq All phosphates are insoluble Exception = s 11. K2CO3(aq) + BaCl2(aq) 2 KCl (aq) + BaCO3 (s) Ions: K + CO3 2 Ba 2+ Cl Group 1A cation = always soluble = aq All carbonate are insoluble Exception = s 12. Na3PO4(aq) + (NH4)2S(aq) NO RXN Ions: Na + PO4 3 NH4 + S 2 Na2S = Group 1A cation = always soluble = aq (NH4)3PO4 = Ammonium cation = always soluble = aq Both aq = No change 13. H3PO4(s) + FeBr3(aq) 3 HBr (aq) + FePO4 (s) Ions: H + PO4 3 Fe 3+ Br Group 1A cation / Acid = always soluble = aq All phosphates are insoluble Exception = s 14. AgNO3(aq) + KCl (aq) KNO3 (aq) + AgCl (s) Ions: Ag + NO3 K + Cl Group 1A cation / Acid = always soluble = aq All chlorides are soluble Ag is an xception = s 15. Na2CO3(aq) + H2SO4 (aq) NO RXN Ions: Na + CO3 2 H + SO4 2 H2CO3 = Group 1A cation / Acid = always soluble = aq Na2SO4 = Group 1A cation = always soluble = aq Both aq = No change 6.D Solutions: Solubility Curves I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS a zero ( 0 ) is better for you than copying Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE. 1. Explain how the solubility of a solution is affected by temperature: In a solid/liquid: as temperature inc, the solubility incr. More energy is available to break IMF s and possibility of breaking due to increased motion In a gas/liquid: solubility of a gas decr as temper incr. More energy is available to overcome IMF s holding gas in solution so it escapes 2. Explain how the solubility of a solution is affected by pressure: In a solid/liquid: no effect. Solids/liquids are not significantly compressible. In a gas/liquid: incr pressure = incr solubility as more gas molecules are forced into solution to release vapor pressure. 13. At 40 C, how many grams of NaNO3 will make a saturated solution if the NaNO3 is added to 100 grams of water? 104 g NaNO3 14. At 80 C, how many grams of KCl can be dissolved in 200 grams of water? 104 g KCl 50. g KCl w/100 g H2O, so x2 for 200g H2O 15. At what temperature will 10 grams of NH3 dissolve completely in 100 grams of water to make a saturated solution? 90. C 16. At 40 C, how many grams of KNO3 can be dissolved in 300 grams of water? 200 g KNO3 62. g KNO3 w/100 g H2O, so x3 for 300g H2O but 300 is 1 sf 17. At 55 C, how many grams of NaNO3 can be dissolved in 50 grams of water? 60 g NaNO g NaNO3 w/40 g H2O, so x½ for 50g H2O and 1 sf Instructions: Using the solubility curve left, answer questions Note while some mental math may be required NW=NC does not apply. 3. Which compound is least soluble at: (A) 20 C? KClO3 (B) 80 C? Ce2(SO4)3 4. Which substance is the most soluble at: (A) 10 C? KI (B) 50 C? KI 5. The solubility of which substance is most affected by changes in temperature? KNO3 as is has the most apparent slope change. 6. The solubility of which substance is least affected by changes in temperature? NaCl as is has the least apparent slope change. 7. Are the following solutions saturated, unsaturated, or supersaturated? (Assume all are dissolved in 100 grams of water.) a. 50 g of NH4Cl at 50 C saturated e. 65 g of NH4Cl at 70 C supersaturated b. 100 g of NaNO3 at 80 C unsaturated f. 30 g of NH3 at 50 C supersaturated c. 30 g of KNO3 at 25 C unsaturated g. 10 g of KClO3 at 20 C supersaturated d. 51 g of KCl at 80 C supersaturated 8. NH3 is a gas. Describe what happens to its solubility as the temperature goes from 20 C to 80 C. solubility decreases as temp increases 9. Which two substances have the same solubility at 58 C? KClO3 & NH3 What is the solubility?24g / 100g H2O 10. Which two substances have the same solubility at 94 C? KCl & NaCl What is the solubility? 53g / 100g H2O 11. For each of the following, indicate the temperature at which the solution described would be saturated. (Assume all are dissolved in 100 grams of water.) Responses should be within +/ 2 C a. 30 grams of NH4Cl 3 C d. 20 grams of KNO3 8 C b. 130 grams of NaNO3 67 C e. 40 grams of KCl 50. C Need the. c. 50 grams of KClO3 92 C f. 60 grams of NH3 15 C 12. For each of these, indicate how many grams of solute (per 100 grams of water) will dissolve. +/ 2g a. NaNO3 at 70 C 132 g NaNO3 c. KI at 20 C 146 g KI b. NH4Cl at 50 C 50. g NH4C Need. d. KClO3 at 90 C 48 g KClO3 18. At 80 C, you have a saturated solution of KClO3. How many grams of solid precipitate will form if the solution is cooled to 50 C? 20 g KClO3 80 = 40. g 50 = 20. g 40.g 20.g = 20. g 19. How many grams of NaNO3 precipitate will form if a saturated solution at 70 C is cooled to 10 C? 51 g NaNO3 70 = 73 g 10 = 22 g 73 g 22 g = 51 g 20. A solution contains 20 g of NH4Cl at 50 C. How many more grams of NH4Cl need to be added to the 100 grams of water for the solution to be saturated? 30. g NH4Cl 50 = 50 g 50. g 20. g = 30. g

7 6.E Solutions: Dilution I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS a zero ( 0 ) is better for you than copying Instructions: ON A SEPARATE SHEET OF PAPER, given the following data, perform the following dilution calculations. Remember, Sig Figs, NW = NC, Boxed Answers and N 3 (this includes labeling and in your set-up!!!) 1. A stock solution of 1.00 M NaCl is available. How many milliliters are needed to make ml of M. M1 V1 = M2 V2 V1 = M2 V2 = M ml M M = 75.0 ml of 1.00 M NaCl 5. A M solution is to be diluted to ml of a M solution. How many ml of the M solution are required? M1 V1 = M2 V2 V1 = M2 V2 = M ml M M = 150 ml of M Solution 2. What volume of M KCl is needed to make ml of M solution? M1 V1 = M2 V2 V1 = M2 V2 = M ml M M = 40.0 ml of M KCl 3. Concentrated H2SO4 is 18.0 M. What volume is needed to make 2.00 L of 1.00 M solution? 6. A stock solution of 10.0 M NaOH is prepared. From this solution, you need to make ml of M solution. How many ml will be required? M1 V1 = M2 V2 V1 = M2 V2 = M ml M M = 9.38 ml of 10.0 M NaOH M1 V1 = M2 V2 V1 = M2 V2 = 1.00 M 2.00 L M M = L of 18.0 M H2SO4 4. Concentrated HCl is 12.0 M. What volume is needed to make 2.00 L of 1.00 M solution? M1 V1 = M2 V2 V1 = M2 V2 = 1.00 M 2.00 L M M = L of M HCl L of M NaNO3 must be prepared from a solution known to be 1.50 M in concentration. How many ml are required? M1 V1 = M2 V2 V1 = M2 V2 = M 2.00 x 10 3 ml M M = 1070 ml of 1.50 M NaNO3 6.F Stoichiometry: Limiting and Excess Reactants I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS a zero ( 0 ) is better for you than copying Instructions: Given the following chemical equations, balance them. Then ON A SEPARATE SHEET OF PAPER, use dimensional analysis to perform the following stoichiometric calculations. You MUST use dimensional analysis to preform each conversion. Remember, Sig Figs, NW = NC, Boxed Answers and N 3 (this includes labeling and in your set-up!!!) 1. Consider the following reaction: 2 Al(s )+ 6 HBr (aq) 2 AlBr3 (aq) + 3 H2 (g) a. When 3.22 moles of Al reacts with 4.96 moles of HBr, how many moles of H2 are formed? Have to compare moles of H2 each reactant will produce Plan: mol Al mol H mol Al 3 mol H2 = 4.83 mol H2 2 mol Al Plan: mol HBr mol H mol HBr 3 mol H2 = 2.48 mol H mol < 4.83 mol 6 mol HBr b. What is the limiting reactant? HBr c. For the reactant in excess, how many moles are left over at the end of the reaction? Have to subtract the mol of Al that HBr reacts with from its initial amount Plan: mol H2 mol H mol HBr 2 mol Al = 1.65 mol Al 3.22 mol Al 1.65 mol Al = 1.57 mol Al 6 mol HBr

8 2. Consider the following reaction: 3 Si (s) + 2 N2 (g) Si3N4 (s) a. When moles of Si reacts with moles of N2, how many moles of Si3N4 are formed? Have to compare moles of Si3N4 each reactant will produce Plan: mol Si mol Si3N4 Plan: mol Si 1 mol Si3N4 = mol Si3N4 mol N2 mol Si3N4 3 mol Si mol N2 1 mol Si3N4 = mol Si3N4 2 mol N2 b. What is the limiting reactant? Si c. For the reactant in excess, how many moles are left over at the end of the reaction? Have to subtract the mol of N2 that Si reacts with from its initial amount Plan: mol Si mol N mol Si 2 mol N2 = mol N mol N mol N2 = 3.33 mol N2 3 mol Si 3. Consider the following reaction: 2 CuCl2 (aq) + 4 KI (aq) 2 CuI (aq) + 4 KCl (aq) + I2 (s) a. When 0.56 moles of CuCl2 reacts with 0.64 moles of KI, how many moles of I2 are formed? Have to compare moles of I2 each reactant will produce. Plan: mol CuCl2 mol I mol CuCl2 1 mol I2 = 0.28 mol I2 Plan: mol KI mol I2 2mol CuCl mol KI 1 mol I2 = 0.16 mol I2 4 mol KI b. What is the limiting reactant? KI c. For the reactant in excess, how many moles are left over at the end of the reaction? Have to subtract the mol of CuCl2 that KI reacts with from its initial amount Plan: mol KI mol CuCl mol KI 2 mol CuCl2 = 0.32 mol CuCl mol CuCl mol CuCl2 = 0.24 mol CuCl2 4 mol KI 4. Consider the following reaction: 4 FeS2 (s) + 11 O2 (g) 2 Fe2O3 (s) + 8 SO2 (g) a. When moles of FeS2 reacts with 5.44 L of O2 (g), how many liters of SO2 (g) are formed? Have to compare L of SO2 each reactant will produce Plan: mol FeS2 L SO mol FeS2 8 mol SO L SO2 (g) = 1193 L SO2 Plan: L O2 L SO2 4 mol FeS2 1 mol SO2 (g) 5.44 L O2 8 L SO2 = 3.96 L SO2 (g) 11 L O2 b. What is the limiting reactant? O2 c. For the reactant in excess, how many moles are left over at the end of the reaction? Have to subtract the mol of FeS2 that O2 reacts with from its initial amount Plan: L O2 mol FeS L O2 1mol O2 (g) 4 mol FeS2 = mol FeS mol FeS mol FeS2 = mol FeS L O2 (g) 11 mol O2

9 5. Consider the following reaction: 3 CaCO3 (aq) + 2 FePO4 (aq) Ca3(PO4)2 (aq) + Fe2(CO3)3 (s) a. When g CaCO3 reacts with g FePO4, how many grams of Fe2(CO3)3 are formed? Have to compare g of Fe2(CO3)3 each reactant will produce Plan: g CaCO3 mol CaCO3 mol Fe2(CO3)3 g Fe2(CO3)3 MM CaCO3 = 1 x g/mol Ca + 1 x g/mol C + 3 x g/mol O = g/mol CaCO3 MM Fe2(CO3)3 = 2 x g/mol Fe + 3 x g/mol C + 9 x g/mol O = g/mol Fe2(CO3) g CaCO3 1 mol CaCO3 1 mol Fe2(CO3) g Fe2(CO3)3 = g Fe2(CO3) g CaCO3 3 mol CaCO3 1 mol Fe2(CO3)3 Plan: g FePO4 mol FePO4 mol Fe2(CO3)3 g Fe2(CO3)3 MM FePO3 = 1 x g/mol Fe + 1 x g/mol P + 4 x g/mol O = g/mol FePO g FePO4 1 mol FePO4 1 mol Fe2(CO3) g Fe2(CO3)3 = g Fe2(CO3) g FePO4 2 mol FePO4 1 mol Fe2(CO3)3 b. What is the limiting reactant? CaCO3 c. For the reactant in excess, how many grams are left over at the end of the reaction? Have to subtract the g of FePO4 that CaCO3 reacts with from its initial amount Plan: g CaCO3 mol CaCO3 mol FePO4 g FePO4 Refere to part A for MM verification data g CaCO3 1 mol CaCO3 2 mol FePO g FePO4 = g FePO g FePO g FePO4 = g FePO4 6. Consider the following reaction: CuCl2 (aq) + 2 NaNO3 (aq) Cu(NO3)2 (aq) + 2 NaCl (aq) g CaCO3 3 mol CaCO3 1 mol FePO4 a. When g CuCl2 reacts with g NaNO3, how many FU of Cu(NO3)2 are formed? Have to compare FU of Cu(NO3)2 each reactant will produce Plan: g CuCl2 mol CuCl2 mol Cu(NO3)2 FU Cu(NO3)2 MM CuCl2 = 1 x g/mol Cu + 2 x g/mol Cl = g/mol CuCl g CuCl2 1 mol CuCl2 1 mol Cu(NO3) x10 23 FU Cu(NO3)2 = x FU Cu(NO3) g CuCl2 1 mol CuCl2 1 mol Cu(NO3)2 Plan: g NaNO3 mol NaNO3 mol Cu(NO3)2 FU Cu(NO3)2 MM NaNO3 = 1 x g/mol Na + 1 x g/mol N + 3 x g/mol O = g/mol NaNO g NaNO3 1 mol NaNO3 1 mol Cu(NO3) x10 23 FU Cu(NO3)2 = x FU Cu(NO3) g NaNO3 2 mol NaNO3 1 mol Cu(NO3)2 b. What is the limiting reactant? NaNO3 c. For the reactant in excess, how many grams are left over at the end of the reaction? Have to subtract the g of CuCl2 that NaNO3 reacts with from its initial amount Plan: g NaNO3 mol NaNO3 mol CuCl2 g CuCl2 Refere to part A for MM verification data g NaNO3 1 mol NaNO3 1 mol CuCl g CuCl2 = g CuCl g NaNO3 2 mol NaNO3 1 mol CuCl g CuCl g CuCl2 = 4.88 g CuCl2 7. Consider the following reaction: 3 FeCl2 (aq) + 2 Na3PO4 (aq) Fe3(PO4)2 (s) + 6 NaCl (aq) a. When 71.3 L of M FeCl2 reacts with 71.3 L of M Na3PO4, how many grams of NaCl are formed? Have to compare g of NaCl each reactant will produce Plan: L M FeCl2 mol FeCl2 mol NaCl g NaCl MM NaCl = 1 x g/mol Na + 1 x g/mol Cl = g/mol NaCl 71.3 L FeCl mol FeCl2 6 mol NaCl g NaCl = 3250 g NaCl 1 L FeCl2 (aq) 3 mol FeCl2 1 mol NaCl Plan: L M Na3PO4 mol Na3PO4 mol NaCl g NaCl MM NaCl = 1 x g/mol Na + 1 x g/mol Cl = g/mol NaCl 71.3 L Na3PO mol Na3PO4 6 mol NaCl g NaCl = 2930 g NaCl 1 L Na3PO4 (aq) 2 mol Na3PO4 1 mol NaCl b. What is the limiting reactant? NaPO4 c. For the reactant in excess, how many FU are left over at the end of the reaction? Have to subtract the L of M FeCl2 that Na3PO4 reacts with from its initial amount and then convert this into FU of FeCl2. Plan: L M Na3PO4 mol Na3PO4 mol FeCl2 L M FeCl L Na3PO mol Na3PO4 3 mol FeCl2 1 L FeCl2 = 64.2 L FeCl2 Plan: L M FeCl2 mol FeCl2 FU FeCl2 1 L Na3PO4 (aq) 2 mol Na3PO mol FeCl2 (aq) 71.3 L FeCl L FeCl2 = 7.10 L FeCl2 (aq) mol FeCl x10 23 FU FeCl2 = 1.67 x FU FeCl2 1 L FeCl2 1 mol FeCl2

10 8. Consider the following reaction: 3 NH4NO3 (aq) + Na3PO4 (aq) (NH4)3PO4 (aq) + 3 NaNO3 (aq) a. When 2.62 L of 2.41 M NH4NO3 reacts with 5.44 L of 4.14 M Na3PO4, how many liters of 2.0M NaNO3 are formed? Have to compare L of 2.0M NaNO3 each reactant will produce Plan: L 2.41 M NH4NO3 mol NH4NO3 mol NaNO3 L 2.0 M NaNO L NH4NO mol NH4NO3 3 mol NaNO3 1 L NaNO3 = 3.16 L of 2.0 M NaNO3 Plan: L 4.14 M Na3PO4 mol Na3PO4 mol NaNO3 L 2.0 M NaNO3 1 L NH4NO3 (aq) 3 mol NH4NO3 2.0 mol NaNO L Na3PO mol Na3PO4 3 mol NaNO3 1 L NaNO3 = 33.8 L of 2.0M NaNO3 1 L Na3PO4 (aq) 1 mol Na3PO4 2.0 mol NaNO3 b. What is the limiting reactant? NH4NO3 c. For the reactant in excess, how many FU are left over at the end of the reaction? Have to subtract the L of 2.0 M NaNO3 that NH4NO3 reacts with from its initial amount and then convert this into FU of Na3PO4. Plan: L 2.41 M NH4NO3 mol NH4NO3 mol Na3PO4 L 4.14 M Na3PO L NH4NO mol NH4NO3 1 mol Na3PO4 1 L Na3PO4 (aq) = L 4.14 M Na3PO4 Plan: L 4.14 M Na3PO4 mol Na3PO4 FU Na3PO4 1 L NH4NO3 (aq) 3 mol NH4NO mol Na3PO L Na3PO LNa3PO4 = 4.93 L Na3PO4 (aq) 4.14 mol Na3PO4 (aq) 6.022x10 23 FU Na3PO4 = 1.23 x FU Na3PO4 1 L Na3PO4 1 mol Na3PO4 6.G Stoichiometry: Theoretical and Percent Yield I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS a zero ( 0 ) is better for you than copying Instructions: Given the following chemical equations, balance them. Then ON A SEPARATE SHEET OF PAPER, use dimensional analysis to perform the following stoichiometric calculations. You MUST use dimensional analysis to preform each conversion before calculating the percent yield. Remember, Sig Figs, NW = NC, Boxed Answers and N 3 (this includes labeling and in your set-up!!!) 1. Given the following equation: K2PtCl4 (aq) + 2 NH3 (s) Pt(NH3)2Cl2 (aq) + 2 KCl (aq) a. Determine the theoretical yield of KCl l if you start with 34.5 grams of NH3. Theoretical yield = perform the stoichiometry as per norm Plan: g NH3 mol NH3 mol KCl g KCl MM NH3 = 1 x g/mol N + 3 x 1.01 g/mol H = g/mol NH3 MM KCl = 1 x g/mol K + 1 x g/mol Cl = g/mol KCl 34.5 g NH3 1 mol NH3 2 mol KCl g KCl = 151 g KCl g NH3 2 mol NH3 1 mol KCl b. Starting with that 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is the percent yield? Have to calculate the theoretical yield and then can compare to the actual citied in the problem. Plan: g NH3 mol NH3 mol Pt(NH3)2Cl2 g Pt(NH3)2Cl2 MM NH3 = 1 x g/mol N + 3 x 1.01 g/mol H = g/mol NH3 MM Pt(NH3)2Cl2 = 1 x g/mol Pt + 2 x g/mol N + 6 x 1.01 g/mol H + 2 x g/mol Cl = g/mol Pt(NH3)2Cl g NH3 1 mol NH3 1 mol Pt(NH3)2Cl g Pt(NH3)2Cl2 = 304 g Pt(NH3)2Cl g NH3 2 mol NH3 1 mol Pt(NH3)2Cl2 % Yield = 76.4 g Pt(NH3)2Cl2 304 g Pt(NH3)2Cl2 x 100% = 25.1 % Yield Pt(NH3)2Cl2 2. Given the following equation: H3PO4 (aq) + 3 KOH (aq) K3PO4 (aq) + 3 H2O (l) a. If 49.0 L of 0.23 M H3PO4 is reacted with excess KOH, determine the percent yield of K3PO4 if you isolate 49.0 g of K3PO4. Have to calculate the theoretical yield and then can compare to the actual citied in the problem. Plan: L 0.23 M H3PO4 mol H3PO4 mol K3PO4 g K3PO4 MM K3PO4 = 3 x g/mol K + 1 x g/mol P + 4 x g/mol O = g/mol K3PO L H3PO mol H3PO4 1 mol K3PO K3PO4 = 2390 g K3PO4 1 L H3PO4 1 mol H3PO4 1 mol K3PO4 % Yield = 49.0 g K3PO g K3PO4 x 100% = 2.05 % Yield K3PO4

11 3. Given the following equation: Al2(SO3)(aq) + 6 NaOH(aq) 3 Na2SO3 (aq) + 3 Al(OH)3 (s) a. If you start with g of Al2(SO3)3 and you isolate g of Na2SO3, what is your percent yield for this reaction? Have to calculate the theoretical yield and then can compare to the actual citied in the problem. Plan: g Al2(SO3)3 mol Al2(SO3)3 mol Na2SO3 g Na2SO3 MM Al2(SO3)3 = 2 x g/mol Al + 3 x g/mol S + 9 x g/mol O = g/mol Al2(SO3)3 MM Na2SO4 = 2 x g/mol Na + 1 x g/mol S + 4 x g/mol O = g/mol Al2(SO3) g Al2(SO3)3 1 mol Al2(SO3)3 3 mol Na2SO g Na2SO3 = g Na2SO g Al2(SO3)3 1 mol Al2(SO3)3 1 mol Na2SO3 % Yield = g Na2SO g Na2SO3 x 100% = % Yield Na2SO3 4. Given the following equation: Al(OH)3 (s) + 3 HCl (aq) AlCl3 (aq) + 3 H2O (l) a. If you start with 50.3 g of Al(OH)3 and you isolate 39.5 g of AlCl3, what is the percent yield? Have to calculate the theoretical yield and then can compare to the actual citied in the problem. Plan: g Al(OH)3 mol Al(OH)3 mol AlCl3 g AlCl3 MM Al(OH)3 = 1 x g/mol Al + 3 x g/mol O + 3 x 1.01 g/mol H = g/mol Al(OH)3 MM AlCl3 = 1 x g/mol Al + 3 x g/mol Cl = g/mol AlCl g Al(OH)3 1 mol Al(OH)3 1 mol AlCl g AlCl3 = 86.0 g AlCl g Al (OH)3 1 mol Al (OH)3 1 mol AlCl3 % Yield = 39.5 g AlCl g AlCl3 x 100% = 45.9 % Yield AlCl3 5. Given the following equation: K2CO3 (s) + 2 HCl (aq) H2O(l) + CO2 (g) + 2 KCl (aq) a. Determine the theoretical yield of KCl if you start with 34.5 g of K2CO3. Theoretical yield = perform the stoichiometry as per norm Plan: g K2CO3 mol K2CO3 mol KCl g KCl MM K2CO3 = 2 x g/mol K + 1 x g/mol C + 3 x g/mol O = g/mol K2CO3 MM KCl = 1 x g/mol K + 1 x g/mol Cl = g/mol KCl 34.5 g K2CO3. 1 mol K2CO3. 2 mol KCl g KCl = 37.2 g KCl g K2CO3. 1 mol K2CO3. 1 mol KCl b. Starting with 34.5 g of K2CO3, and you isolate 3.4 g of H2O, what is the percent yield? Have to calculate the theoretical yield and then can compare to the actual citied in the problem. Plan: g K2CO3 mol K2CO3 mol H2O g H2O MM K2CO3 = 2 x g/mol K + 1 x g/mol C + 3 x g/mol O = g/mol K2CO3 MM H2O = 2 x 1.01 g/mol H + 1 x g/mol O = g/mol H2O 34.5 g K2CO3. 1 mol K2CO3. 1 mol H2O g H2O = 4.50 g H2O g K2CO3. 1 mol K2CO3. 1 mol H2O % Yield = 3.4 g H2O 4.50 g H2O x 100% = 76 % Yield H2O 6. Given the following equation: H2SO4 (aq) + Ba(OH)2 (aq) BaSO4 (s) + 2 H2O (l) a. If L of 0.50 M H2SO4 is reacted with excess Ba(OH)2, determine the percent yield of BaSO4 if you isolate g of BaSO4. Have to calculate the theoretical yield and then can compare to the actual citied in the problem. Plan: L 0.50 M H2SO4 mol H2SO4 mol BaSO4 g BaSO4 MM BaSO4 = 1 x g/mol Ba + 1 x g/mol S + 4 x g/mol O = g/mol BaSO L H2SO mol H2SO4 1 mol BaSO g BaSO4 = g BaSO4 1 L H2SO4 1 mol H2SO4 1 mol BaSO4 % Yield = g BaSO g BaSO4 x 100% = % Yield BaSO4 7.Given the following equation: 3 CaCl2 (s) + 2 Li3PO4 (aq) 6 LiCl (aq) + Ca3(PO4)2 (s) a. If you start with 82.4 g of CaCl2 and you isolate 82.4 g of Ca3(PO4)2, what is your percent yield for this reaction? Have to calculate the theoretical yield and then can compare to the actual citied in the problem. Plan: g CaCl2 mol CaCl2 mol Ca3(PO4)2 g Ca3(PO4)2 MM CaCl2 = 1 x g/mol Ca + 2 x g/mol Cl = g/mol CaCl2 MM Ca3(PO4)2 = 3 x g/mol Ca + 2 x g/mol P + 8 x g/mol O = g/mol Ca3(PO4) g CaCl2 1 mol CaCl2 1 mol Ca3(PO4) g Ca3(PO4)2 = 76.8 g Ca3(PO4) g CaCl2 3 mol CaCl2 1 mol Ca3(PO4)2 % Yield = 82.4 g Ca3(PO4) g Ca3(PO4)2 x 100% = 107 % Yield Ca3(PO4)2

12 8. Given the following equation: Cr(OH)3 (s) + 3 HI (aq) CrI3 (aq) + 3 H2O (l) a. If you start with 50.3 g of Cr(OH)3 and you isolate 39.5 g of CrI3, what is the percent yield? Have to calculate the theoretical yield and then can compare to the actual citied in the problem. Plan: g Cr(OH)3 mol Cr(OH)3 mol CrI3 g CrI3 MM Cr(OH)3 = 1 x g/mol Cr + 3 x g/mol O + 3 x 1.01 g/mol H = g/mol Cr(OH)3 MM CrI3 = 1 x g/mol Cr + 3 x g/mol I = g/mol CrI g Cr(OH)3 1 mol Cr(OH)3 1 mol CrI g CrI3 = 211 g CrI g Cr(OH)3 1 mol Cr(OH)3 1 mol CrI3 % Yield = 39.5 g CrI3 211 g CrI3 x 100% = 18.7 % Yield CrI3 6.H Stoichiometry: Review I Pledge : ( Initial ) DON T FORGET WHAT THIS REPRESENTS a zero ( 0 ) is better for you than copying Instructions: ON A SEPARATE SHEET OF PAPER, use dimensional analysis to perform the following stoichiometric calculations. You MUST use dimensional analysis to preform each stoichiometric conversion. Remember, Sig Figs, NW = NC, Boxed Answers and N 3 (this includes labeling and in your set-up!!!) 1. Chemical Equation: 2 Na3PO4 (aq) + 3 Ba(NO3)2 (aq) Ba3(PO4)2 (ppt) + 6 NaNO3 (aq) If you react 5.7 moles of Na3PO4 with barium nitrate. Notice that in the problem you are only given one reactant amount and an unbalanced chemical equation. Thus you can assume that you have all that you need of the other reactant and you do NOT need to worry about limiting reactant calculations. Thus the 5.7 moles of Na3PO4 is the limiting reactant. Also you should notice that you are given the amounts in moles, so there is no reason to change to moles, you are already in moles! a. How many moles of barium nitrate would you need to go with it? Plan: mol Na3PO4 mol Ba(NO3)2 5.7 mol Na3PO4 3 mol Ba(NO3)2 = 8.6 mol Ba(NO3)2 2 mol Na3PO4 b. What mass of sodium nitrate would you be able to produce? Plan: mol Na3PO4 mol NaNO3 g NaNO3 MM NaNO3 = 1 x g/mol Na + 1 x g/mol N + 3 x g/mol O = g/mol NaNO3 5.7 mol Na3PO4 6 mol NaNO g NaNO3 = 1500 g NaNO3 2 mol Na3PO4 1 mol NaNO3 2. Chemical Equation: 2 Mg + O2 2 MgO If you have 20. moles of oxygen gas and 10. moles of magnesium. Because you are given amounts of both reactants, this is a limiting reactant problem with an unbalanced chemical equation provided. You must first determine which reactant limits. Be sure and note that the amounts are already in moles, so there is no need to change to moles! a. Determine which substance limits the reaction. Plan: mol O2 mol MgO 20. mol O2 2 mol MgO = 40. mol Mg 1 mol O2 Plan: mol Mg mol MgO 10. mol Mg 2 mol MgO = 10. mol Mg Mg is limiting 2 mol Mg b. Is there anything left over? Which substance? How many moles? Have to determine how much O2 will react, as the previous problem determined it was in excess, with 10. mol of Mg and then subtract it from its initial amount Plan: mol O2 mol MgO 10. mol Mg 1 mol O2 = 5.0 mol O2 20. mol O2 5.0 mol O2 = 15 mol O2 is left over 2 mol Mg c. Keep your wits about you, read the question.you can do this one! If you wanted to use up all the excess reactant, how many more moles of the limiting reactant would you need to add? Have to determine how much Mg is needed to react with the left over O2. With 15 mol of O2 left, the stoichiometry should be pretty straight forward at this point. Plan: mol O2 mol Mg 15 mol O2 2 mol Mg = 30. mol Mg more 1 mol O2

13 3. Balanced Equation: 2 H2 + O2 2 H2O You have 50.0 grams of O2 and 10.0 grams of H2 to react together. Because you are given two reactants, this is clearly a limiting reactant problem. a. Which gas is the limiting reactant? Plan: g O2 mol O2 mol H2O MM O2 = 2 x g/mol O = g/mol O g O2 1 mol O2 2 mol H2O = 3.13 mol O g O2 1 mol O2 Plan: g H2 mol H2 mol H2O MM O2 = 2 x 1.01 g/mol O = 2.02 g/mol H g H2 1 mol H2 2 mol H2O = 4.95 mol H2O O2 is limiting 2.02 g H2 2 mol H2 b. Which gas is left over? How many grams of it? If O2 is limiting, then H2 is in excess. Have to subtract the g of H2 that O2 reacts with from its initial amount. Plan: g O2 mol O2 mol H2 g H2 Refere to part A for MM verification data 50.0 g O2 1 mol O2 2 mol H g H2 = 6.31 g H2 are consumed g O2 1 mol O2 1 mol H g H g H2 = 3.7 g H2 remain 4. Balanced Equation: 2 Al + 3 H2SO4 Al2(SO4)3 + 3 H2 You combine 20.0 g of aluminum and 30.0 g of sulfuric acid. Again, this is a limiting reactant problem because you are given mass values for both reactants. Determine the mass of one product. This will allow you to save time / work for the second as you will be able to identify the limiting reactant from this initial calc. NOTE, both product amounts must be calculated in this instance. a. How many grams of each product could you produce? Plan: g Al mol Al mol Al2(SO4)3 g Al2(SO4)3 MM Al2(SO4)3 = 2 x g/mol Al + 3 x g/mol S + 12 x g/mol O = g/mol Al2(SO4) g Al 1 mol Al 1 mol Al2(SO4) g Al2(SO4)3 = 127 g Al2(SO4) g Al 2 mol Al 1 mol Al2(SO4)3 Plan: g H2SO4 mol H2SO4 mol Al2(SO4)3 g Al2(SO4)3 MM H2SO4 = 2 x 1.01 g/mol H + 1 x g/mol S + 4 x g/mol O = g/mol H2SO g H2SO4 1 mol H2SO4 1 mol Al2(SO4) g Al2(SO4)3 = 34.9 g Al2(SO4) g H2SO4 3 mol H2SO4 1 mol Al2(SO4)3 As you have now determined that H2SO4 is limiting, you really only need to carry through it s stoichiometric conversion to H2. However, I have shown both for those of you that chose to convert to H2 rather than Al2(SO4)3 first. Plan: g H2SO4 mol H2SO4 mol H3 g H2 MM H2SO4 = 2 x 1.01 g/mol H + 1 x g/mol S + 4 x g/mol O = g/mol H2SO4 MM H2 = 2 x 1.01 g/mol H = 2.02 g/mol H g H2SO4 1 mol H2SO4 3 mol H g H2 = g H g H2SO4 3 mol H2SO4 1 mol H2 Plan: g Al mol Al mol Al2(SO4)3 g Al2(SO4) g Al 1 mol Al 3 mol H g H2 = 2.25 g H g Al 2 mol Al 1 mol H2 5. Balanced Equation: Mg3N2 + 6 H2O 3 Mg(OH)2 + 2 NH3 In the lab you did this reaction with 58.0 g of magnesium nitride and 62.1 g of water, and you produced 10.0 g of ammonia. Again, this is a limiting reactant problem because you are given mass values for both reactants. a. Determine the % yield of ammonia. Have to calculate the theoretical yield based on the unidentified limiting reactant. Can then can compare to the actual citied in the problem. Plan: g Mg3N2 mol Mg3N2 mol NH3 g NH3 MM Mg3N2 = 3 x g/mol Mg + 2 x g/mol N = g/mol Mg3N2 MM NH3 = 1 x g/mol N + 3 x 1.01 g/mol H = g/mol NH g Mg3N2 1 mol Mg3N2 2 mol NH g NH3 = 18.1 g NH3 = Limited Theo g Mg3N2 1 mol Mg3N2 1 mol NH3 Plan: g H2O mol H2O mol NH3 g NH3 MM H2O = 2 x 1.01 g/mol H + 1 x g/mol O = g/mol H2O MM NH3 = 1 x g/mol N + 3 x 1.01 g/mol H = g/mol NH g H2O 1 mol H2O 2 mol NH g NH3 = 19.6 g NH g H2O 6 mol H2O 1 mol NH3 % Yield = 10.0 g NH g NH3 x 100% = 55.2 % Yield NH3

14 b. Is there any reactant left over from the actual reaction? What masses? Knowing that only 10.0 g of NH3 are produced, you must go back and find how much of each react was consumed to produce this amount. Then you may subtract these values from the initial amount of reacts present. Plan: g NH3 mol NH3 mol Mg3N2 g Mg3N2 MM Mg3N2 = 3 x g/mol Mg + 2 x g/mol N = g/mol Mg3N2 MM NH3 = 1 x g/mol N + 3 x 1.01 g/mol H = g/mol NH g NH3 1 mol NH3 1 mol Mg3N g Mg3N2 = 32.0 g Mg3N g NH3 2 mol NH3 1 mol Mg3N g Mg3N g Mg3N2= 26.0 g Mg3N2 remain Plan: g NH3 mol NH3 mol H2O g H2O MM H2O = 2 x 1.01 g/mol H + 1 x g/mol O = g/mol H2O MM NH3 = 1 x g/mol N + 3 x 1.01 g/mol H = g/mol NH g NH3 1 mol NH3 6 mol H2O g H2O = 31.7 g H2O g NH3 2 mol NH3 1 mol H2O 62.1 g H2O 31.7 g H2O = 30.4 g H2O remain 6. Balanced Equation: Au2S3 + 3 H2 3 H2S + 2 Au In the lab, 505 g of gold(iii) sulfide is reacted with 5.00 g of hydrogen gas and 327 g of gold was produced. Read the problem carefully to note that you are given two reactant masses which means you must determine which one limits. a. Which reactant is limits the reaction? As the problem does not ask you to go to any specific unit, just to find the reactant that limits, you only have to convert to moles of either product to compare and determine its ID. Plan: g Au2S3 mol Au2S3 mol Au MM Au2S3 = 2 x g/mol Al + 3 x g/mol S = g/mol Au2S3 505 g Au2S3 1 mol Au2S3 2 mol Au = 2.06 mol Au g Au2S3 1 mol Au2S3 Plan: g H2 mol H2 mol Au MM H2 = 2 x 1.01 g/mol H = 2.02 g/mol H g H2 1 mol H2 2 mol Au = 1.65 mol Au H2 is limiting 2.02 g H2 3 mol H2 b. Which reactant is in excess theoretically? What mass of the excess is left over? If H2 is limiting, then Au2S3 is in excess. Have to subtract the g of Au2S3 that H2 reacts with from its initial amount. Plan: g H2 mol H2 mol Au2S3 g Au2S3 See part a. for MM verification 5.00 g H2 1 mol H2 1 mol Au2S g Au2S3 = 404 g Au2S g H2 3 mol H2 1 mol Au2S3 505 g Au2S3 404 g Au2S3 = 101 g Au2S3 are left over c. Determine the % yield of gold. What does this % yield tell you about the purity of your gold? This is the pitfall of only going to moles in a. We now have to determine the number of grams of Au that 5.00 g H2 would theoretically produce to calculate the percent yield. Plan: g H2 mol H2 mol Au g Au See part a. for MM verification 5.00 g H2 1 mol H2 2 mol Au g Au = 325 g Au 2.02 g H2 3 mol H2 1 mol Au % Yield = 327 g Au 325 g Au x 100% = 101 % Yield Au 7. Balanced Equation: 2 C3H6 + 2 NH3 + 3 O2 2 C3H3N + 6 H2O You have access to 12 moles of ethene and 15 moles of ammonia and 16 moles of oxygen. In this problem, like our S mores problem coming up in lab, there are three reactants. One of them will limit. Note that the amounts are given in moles thus there is no need to change to moles. a. What mass of water can be produced? I chose to convert to H2O, it is perfectly valid if you converted to C3H3N Plan: mol C3H6 mol H2O 12 mol C3H6 6 mol H2O = 36 mol O2 Plan: mol NH3 mol H2O 2 mol C3H6 15 mol NH3 6 mol H2O = 45 mol O2 2 mol NH3 Plan: mol O2 mol H2O 16 mol O2 6 mol H2O = 32 mol O2 O2 is limiting 3 mol O2

15 8. Balanced Equation: 2 Al + 2 H3PO4 2 AlPO4 + 3 H2 You would like to produce 5.00 g of aluminum phosphate. Use the amount of aluminum phosphate that you want to produce to figure out how much aluminum that you must start with. a. What mass of aluminum should you start with to react with your excess mass of phosphoric acid? Plan: g H3PO4 mol H3PO4 mol Al g Al MM H3PO4 = 3 x 1.01 g/mol H + 1 x g/mol P + 4 x g/mol O = g/mol H3PO g Al3PO4 1 mol H3PO4 2 mol Al g Al = 1.38 g Al g H3PO4 2 mol H3PO4 1 mol Al b. How many moles of hydrogen gas did you produce at the same time? Plan: g H3PO4 mol H3PO4 mol H2 See part a. for MM verification work 5.00 g Al3PO4 1 mol H3PO4 3 mol H2 = mol H g H3PO4 2 mol H3PO4 9. Balanced Equation: 4 Al + 3 O2 2 Al2O3 When reacting aluminum with excess oxygen, the reaction consistently gives only a 75% yield. What mass of aluminum should you start with if you would like to produce 10.0 g of aluminum oxide in the lab? The only trick in this problem is to be sure and deal with the 75% yield. This means that you have to use an amount of Al that would (in theory) produce more Al2O3, but because of the yield, you would only end up with 10 g. Plan: determine what amount of Al2O3 would have been produced if the rxn was 100% Convert 75% to decimal form of 0.75, noting too that it is only 2 sig figs 10.0 g Al2O = 13 g Al2O3 should have been produced Plan: g Al2O3 mol Al2O3 mol Al mol Al MM Al2O3 = 2 x g/mol Al + 3 x g/mol O = g/mol H3PO4 13 g Al2O3 1 mol Al2O3 4 mol Al g Al = 6.9 g Al g Al2O3 2 mol Al2O3 1 mol Al REMEMBER.PLEASE TREAT HOMEWORK AS PRACTICE COPYING WILL NOT HELP YOU.