2. What is the molarity of g of H2SO4 dissolved in 1.00 L of solution? g H2SO4 1 mol H2SO4 = M H2SO4 1 L 98.

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1 8.A Stoichiometry: Molarity 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? 28.0 g NaCl 1 mol NaCl = M NaCl3 1 L g NaCl 11. Determine the number of moles of solute to prepare these solutions: a liters of a 2.00 M Cu(NO3)2 solution L Cu(NO3) mol Cu(NO3)2 = 4.70 mol Cu(NO3)2 1 L Cu(NO3)2 2. What is the molarity of g of H2SO4 dissolved in 1.00 L of solution? 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? 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? 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? 10.0 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? 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? 2.45 g H2SO4 1 mol H2SO4 1 L H2SO ml g H2SO mol H2SO4 1 L = 1.39 ml H2SO4 8. What volume (in ml) of 12.0 M HCl is needed to contain 3.00 moles of HCl? 3.00 mol HCl 1 L H2SO ml 12.0 mol HCl 1 L = 250. ml HCl 9. What mass of KCl is there in 2.50 liters of 0.50 M KCl solution? 2.50 L KCl 0.50 mol KCl g KCl 1 L KCl 1 mol KCl = 93.2 g KCl 10. Determine the molarity of these solutions: a moles of Li2SO3 dissolved to make 2.04 liters of solution mol Li2SO3 = 2.29 M Li2SO L b moles of Al2O3 to make liters of solution mol Al2O3 = M Al2O L c grams of Na2CO3 to make liters of solution g Na2CO3 1 mol Na2CO3 = M Na2CO L g Na2CO3 d grams of (NH4)2CO3 to make 250 ml of solution g (NH4)2CO3 1 mol (NH4)2CO3 = M (NH4)2CO L g (NH4)2CO3 b ml of a molar Pb(NO3)2 solution L Pb(NO3) mol Pb(NO3)2 = mol Pb(NO3)2 1 L Pb(NO3)2 c L of a M MgCO3 solution L MgCO mol MgCO3 = 1.50 mol MgCO3 1 L MgCO3 d L of a 3.76-molar Na2O solution L Na2O 3.76 mol Na2O = 23.3 mol Na2O 1 L Na2O 12. Determine the grams of solute to prepare these solutions: a liters of a M Cu(NO3)2 solution 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 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 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 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 mol Li2SO3 1 L Li2SO3 = 1.20 L Li2SO mol Li2SO3 b moles of Al2O3 to make a M solution mol Al2O3 1 L Al2O3 = L Li2SO mol Al2O3 c grams of Na2CO3 to make a M solution g Na2CO3 1 mol Na2CO3 1 L Na2CO3 _ g Na2CO mol Na2CO3 = L Na2CO3 d grams of (NH4)2CO3 to make a molar solution g (NH4) 2CO3 1 mol (NH4) 2CO3 1 L (NH4) 2CO g (NH4) 2CO mol (NH4) 2CO3 = L (NH4)2CO3

2 8.B Stoichiometry: Mole Map Basics 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 F mol F L F2 (g) = 7.2 L F2 (g) b L of 2.0 M NaCl g NaCl 1 mol F L NaCl 2.0 mol NaCl(aq) g NaCl = 656 g NaCl c g Cl2 L Cl2 1 L NaCl 1 mol NaCl g Cl2 1 mol Cl L Cl2 (g) = L Cl2 (g) g Cl2 1 mol Cl2 2. Pb(OH)2(aq)+ _2_ HCl(aq) _2_ H2O(l) + PbCl2 (s) a mol Pb(OH)2 mol of H2O 2.42 mol Pb(OH)2 2 mol H2O = 4.84 mol H2O b. 120 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) x 10 7 mol PbCl2 1 mol Pb(OH)2 = 2.02 x 10 7 mol Pb(OH)22. 1 mol PbCl2 3. _2_ AlBr3 (aq) + _3_ K2SO4 (aq) _6_ KBr (aq) + Al2(SO4)3(s) a g K2SO4 g of 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) 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 K2SO g KBr 1 mol KBr 3 mol K2SO g K2SO4 = 5.51x 10 4 g K2SO g KBr 6 mol KBr 1 mol K2SO4 4. _2_ C7H10 (s) + _19_ O2 (g) _14_ CO2 (g) + _10_ H2O (g) a L O2 L CO L O2 14 L CO2 = L CO2 19 L O2 b L H2O mol C7H L H2O 1 mol H2O 2 mol C7H10 = mol C7H L of H2O(g) 10 mol H2O c L CO2 g C7H L CO2 1 mol CO2 2 mol C7H g C7H10 = 3.37 g C7H L of CO2 (g) 14 mol CO2 1 mol C7H10 5. FeCl3 (aq) + _3_ NaOH (aq) Fe(OH)3(s) + _3_ NaCl(aq) a g NaCl g of Fe(OH) g NaCl 1 mol NaCl 1 mol Fe(OH) g Fe(OH)3 = g Fe(OH) g NaCl 3 mol NaCl 1 mol Fe(OH)3 b mol Fe(OH)3 L of 1.24 M NaCl 9.65 mol Fe(OH)3 3 mol NaCl (aq) 1 L of NaCl (aq) = 23.3 L NaCl 1 mol Fe(OH) mol NaCl c L of 4.5 M NaOH L of 6.0 M FeCl L NaOH 4.5 mol NaOH 1 mol FeCl3 1 L FeCl3 = L FeCl33 1L NaOH 3 mol NaOH 6.0 mol FeCl3

3 6. _2_ Ag2O(s) _4_ Ag(s) + O2(g) a L O2 g Ag 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 O g Ag2O 1 mol Ag2O 1 mol O L O2 = g Fe(OH) g Ag2O 2 mol Ag2O 1 mol O2 c g Ag FU Ag2O 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 MgBr x atoms K 1 FU MgBr2 = 7.10 x FU MgBr2 2 atoms K b L of 0.53 M MgBr2 Atoms Mg 5.24 L MgBr mol MgBr2 1 mol Mg 6.022x10 23 atoms Mg = 1.67 x atoms Mg 1 L MgBr2 1 mol MgBr2 1 mol Mg c. 8.8 x FU KBr g K 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 CaCl L HCl 0.53 mol HCl 1 mol CaCl g CaCl2 = 13 L HCl 1 L HCl 2 mol HCl 1 mol CaCl2 b x FU CaCO3 L of M CaCl x FU CaCO3 1 mol CaCO3 1 mol CaCl2 1 L CaCl2 = 17.9 L CaCl x10 23 FU CaCO3 1 mol CaCO mol CaCl2 c g CaCl2 FU H2CO 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 9. _2_ NaBr (aq) + CaF2 (aq) _2_ NaF(aq) + CaBr2(s) a g CaF2 g NaF 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 CaBr x FU NaBr 1 FU CaBr2 = 2.56 x FU CaBr2 2 FU NaBr c mol NaF FU CaF 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 HNO L H2SO mol H2SO4 2 mol HNO g HNO2 = 25.2 g HNO2 1 L H2SO4 1 mol H2SO4 1 mol HNO2 b. 930 mol NaNO2 g H2SO4 930 mol NaNO2 1 mol H2SO g H2SO4 = g H2SO4 2 mol NaNO2 1 mol H2SO4 c x FU HNO2 L of 0.31 M Na2SO x FU HNO2 1 mol HNO2 1 mol Na2SO4 1 L Na2SO4 = 2.57x10 8 L Na2SO x10 23 FU HNO2 2 mol HNO mol Na2SO4

4 8.C Stoichiometry: Limiting and Excess Reactants 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? 3.22 mol Al 3 mol H2 = 4.83 mol H2 2 mol Al 4.96 mol HBr 3 mol H2 = 2.48 mol H2 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? 4.96 mol HBr 2 mol Al = 1.65 mol Al 3.22 mol Al 1.65 mol Al = 1.57 mol Al 6 mol HBr 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? mol Si 1 mol Si3N4 = 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? 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? 0.56 mol CuCl2 1 mol I2 = 0.28 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? 0.64 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? mol FeS2 8 mol SO L SO2 (g) = 1193 L SO2 4 mol FeS2 1 mol SO2 (g) 5.44 L O2 8 L SO2 = 3.96 L SO2 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? 5.44 L O2 1mol O2 (g) 4 mol FeS2 = mol FeS mol FeS mol FeS2 = mol FeS L O2 (g) 11 mol O2

5 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? 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) 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? g CaCO3 1 mol CaCO3 2 mol FePO g FePO4 = g FePO g FePO g FePO4 = g FePO g CaCO3 3 mol CaCO3 1 mol FePO4 6. Consider the following reaction: CuCl2 (aq) + _2_ NaNO3 (aq) Cu(NO3)2 (aq) + _2_ NaCl (aq) a. When g CuCl2 reacts with g NaNO3, how many FU of Cu(NO3)2 are formed? 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) 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? 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? 71.3 L FeCl mol FeCl2 6 mol NaCl g NaCl = 3250 g NaCl 1 L FeCl2 (aq) 3 mol FeCl2 1 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? 71.3 L Na3PO mol Na3PO4 3 mol FeCl2 1 L FeCl2 = 64.2 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 8. Consider the following reaction: 1 L FeCl2 1 mol FeCl2 _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? 2.62 L NH4NO mol NH4NO3 3 mol NaNO3 1 L NaNO3 = 3.16 L 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 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? 2.62 L NH4NO mol NH4NO3 1 mol Na3PO4 1 L Na3PO4 (aq) = L 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 8.D Stoichiometry: Theoretical and Percent Yield 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 NH 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? 34.5 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 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 3. Given the following equation: Al2(SO3)(aq) + _6_ NaOH(aq) _3_ Na2SO3 (aq) + _2_ 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? 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? 50.3 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 K2CO 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? 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 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:

7 _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? 82.4 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 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? 50.3 g Cr(OH)3 1 mol Cr(OH)3 1 mol CrI g CrI3 = 236 g CrI g Cr(OH)3 1 mol Cr(OH)3 1 mol CrI3 % Yield = 39.5 g CrI3 236 g CrI3 x 100% = 16.7 % Yield CrI3

8 8.E Stoichiometry: Review 1. Notice that in the problem you are only given one reactant amount, 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. 8.6 moles of barium nitrate can be formed b g sodium nitrate can be formed (Notice that this answer has been rounded to 2 significant figures.) 2. Because you are given amounts of both reactants, this is a limiting reactant problem. 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. Mg limits b. oxygen is left over. Thus you must calculate the amount of O2 needed to go with the Mg. 15 moles O left over c. This time you must calculate how much magnesium you would need to go with all the oxygen that you were given. Since you already started with 10 moles of Mg, you would need 30 moles of Mg more to use up all the oxygen gas 3. Because you are given two reactants, this is clearly a limiting reactant problem. a. Oxygen gas limits b g H2 left over 4. Again, this is a limiting reactant problem because you are given mass values for both reactants. a. H2SO4 limits = 34.9 g Al2(SO4)3 = g H2 can be produced 5. Again, this is a limiting reactant problem because you are given mass values for both reactants. a. These two numbers are so close that it is telling you that you have just the right proportions of each. b. You could use either value to determine the theoretical value of ammonia, because both reactants would run out at the same time and there would not be any reactant left over. part a continued: 51.3 % yield of ammonia 6. Read the problem carefully to note that you are given two reactant masses which means you must determine which one limits. a. hydrogen limits b. Since hydrogen limits. = 101 g of Au2S3 are left over. c. 99 % yield. 7. In this problem, like our S mores problem, 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. oxygen limits = 576 g of water can be produced 8. Use the amount of aluminum phosphate that you want to produce to figure out how much aluminum that you must start with. a g of Aluminum needed. b mole of H2 will be produced as well. 9. 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. a. (10 g experimental Al2O3 / x g theoretical) * 100 = 75 %, Solve to find out that you need to theoretically produce 13.3 g of Al2O3 = 7.04 g of Al should be started with.

9 8.F Solutions: The Basics 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 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, 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 s = ionic compounds that dissociate and as a result are able to conduct electricity Nonelectroltes are molecular compounds that dissolve in water, but as they are not ionic and thus cannont dissociate, they are unable to conduct electrity. 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 changes 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. 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 interations = greater effect. Instructions: Given the following reactants, write a complete and balanced chemical equation for the double replacment reaction using your references. 10. _3_ Ca(OH)2(aq) + 2 _2_ H3PO4(aq) 6 H2O (l_) + Ca3(PO4)2 (s) 11. K2CO3(aq) + BaCl2(aq) 2 KCl (laq) + BaCO3 (s) 12. Na3PO4(aq) + (NH4)2S(aq) NO RXN as Na2S (aq) + (NH4)3PO4 (aq) 13. H3PO4(s) + FeBr3(aq) 3 HBr (aq) + FePO4 (s) 14. AgNO3(aq) + KCl (aq) AgCl (s) + KNO3 (aq) 15. Na2CO3(aq) + H2SO4 (aq) NO RXN as Na2SO4 (aq) + H2CO3 (aq)

10 8.G Solutions: Solubility Curves 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 solbuitly as more gas molecules are forced into solouiton to releave 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

11 8.H Solutions: Dilution 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. 5. A M solution is to be diluted to ml of a M solution. How many ml of the M solution are required? V1 = M2 V2 V1 = M2 V2 = M ml 1.00 M = 75.0 ml of 1.00 M NaCl V1 = M2 V2 V1 = M2 V2 = M ml M = 150 ml of M Solution 2. What volume of M KCl is needed to make ml of M solution? V1 = M2 V2 V1 = M2 V2 = M ml M = 40.0 ml of M KCl 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? V1 = M2 V2 V1 = M2 V2 = M ml 10.0 M = 9.38 ml of 10.0 M NaOH 3. Concentrated H2SO4 is 18.0 M. What volume is needed to make 2.00 L of 1.00 M solution? V1 = M2 V2 V1 = M2 V2 = 1.00 M 2.00 L 18.0 M = L of 18.0 M H2SO L of M NaNO3 must be prepared from a solution known to be 1.50 M in concentration. How many ml are required? V1 = M2 V2 V1 = M2 V2 = M 2.00 x 10 3 ml 1.50 M = 1070 ml of 1.50 M NaNO3 4. Concentrated HCl is 12.0 M. What volume is needed to make 2.00 L of 1.00 M solution? V1 = M2 V2 V1 = M2 V2 = 1.00 M 2.00 L 12.0 M = L of M HCl