18. Cobalt (III) carbonate C02(CO3)3 Iron (III) phosphide FeP

45. C + O O  C O O

46. C + O  C O O

47. C + O  C O O O C

48. C + O  C O O O C C

49. H 2 +H2OH2OO2O2  Make a table to keep track of where you are at

50. Example H 2 +H2OH2OO2O2  Need twice as much O in the product RP H O 2 2 2 1

51. Example H 2 +H2OH2OO2O2  Changes the O RP H O 2 2 2 1 2

52. Example H 2 +H2OH2OO2O2  Also changes the H RP H O 2 2 2 1 2 2

53. Example H 2 +H2OH2OO2O2  Need twice as much H in the reactant RP H O 2 2 2 1 2 2 4

54. Example H 2 +H2OH2OO2O2  Recount RP H O 2 2 2 1 2 2 4 2

55. Example H 2 +H2OH2OO2O2  The equation is balanced, has the same number of each kind of atom on both sides RP H O 2 2 2 1 2 2 4 2 4

56. Example H 2 +H2OH2OO2O2  This is the answer RP H O 2 2 2 1 2 2 4 2 4 Not this

58. #4

64. If you don’t know the answer to this—look it up!!

65. Due Today: Balancing equations worksheet Homework Due Monday Chapter 10-1-CORNELL notes and questions #1-4 Through “Mass-Mole relationships” in packet NO WASTING TIME TODAY

107. 106 Stoichiometry Ratios are found within a chemical equation. 2HCl + Ba(OH) 2  2H 2 O + BaCl 2 1 1 2 moles of HCl react with 1 mole of Ba(OH) 2 to yield 2 moles of H 2 O and 1 mole of BaCl 2 coefficients give MOLAR RATIOS

108. 107 When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of NO 2 can be produced from 4.3 moles of N 2 O 5 ? = moles NO 2 4.3 mol N2O5N2O5 8.6 b. How many moles of O 2 can be produced from 4.3 moles of N 2 O 5 ? = mole O 2 4.3 mol N2O5N2O5 2.2 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol? mol 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol ? mol Mole – Mole Conversions Units match

110. 3.34 moles Al 2 O 3 2 moles Al 2 O 3 3 mole O 2 =5.01 moles O 2

114. Periodic Table Moles A Moles B Mass g B Periodic Table Balanced Equation Mass g A Decide where to start based on the units you are given and stop based on what unit you are asked for

115. 114 When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of N 2 O 5 were used if 210g of NO 2 were produced? = moles N 2 O 5 210 g NO 2 2.28 b. How many grams of N 2 O 5 are needed to produce 75.0 grams of O 2 ? = grams N 2 O 5 75.0 g O2O2 506 gram ↔ mole and gram ↔ gram conversions 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 210g? moles 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 75.0 g ? grams Units match

116. 115 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? First write a balanced equation. Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 Gram to Gram Conversions

117. 116 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 Now let’s get organized. Write the information below the substances. 3.45 g ? grams Gram to Gram Conversions

118. 117 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 3.45 g ? grams Let’s work the problem. = g AlCl 3 3.45 g Al We must always convert to moles.Now use the molar ratio.Now use the molar mass to convert to grams. 17.0 Units match gram to gram conversions

120. Calculate this as 2 problems. Follow the steps exactly as we have done each time, but do it TWICE. Whichever one will produce the smaller amount is the limiting reactant Easy, right?

129. 6.45 g H 2 O 18.02 g H 2 O 1 mol H 2 O 2 mol H 2 O 1 mol O 2 22.4 L O 2

131. 17.5 L O2O2 22.4 L O2O2 1 mol O2O2 2 O2O2 1 CH 4 1 mol CH 4 22.4 L CH 4 = 8.75 L CH 4 22.4 L O 2 1 mol O 2 1 mol CH 4 22.4 L CH 4

133. 17.5 L CH 4 1 L CH 4 1 L CO 2 = 17.5 L CO 2

136. Camels store the fat tristearin (C 57 H 110 O 6 ) in the hump. As well as being a source of energy, the fat is a source of water, because when it is used the reaction takes place. What mass of water can be made from 1.0 kg of fat? X g H 2 O = 1 kg ‘fat” (1000 g ‘fat’) (1 mol “fat”) (110 mol H 2 O) (18 g H 2 O) (1 kg ‘fat’) (890 g ‘fat’) (2 mol ‘fat’) (1 mol H 2 O) X = 1112 g H 2 O or 1.112 liters water 2 C 57 H 110 O 6 (s) + 163 O 2 (g)  114 CO 2 (g) + 110 H 2 O(l)

137. The compound diborane (B 2 H 6 ) was at one time considered for use as a rocket fuel. How many grams of liquid oxygen would a rocket have to carry to burn 10 kg of diborane completely? (The products are B 2 O 3 and H 2 O). B 2 H 6 + 3 O 2  B 2 O 3 + 3 H 2 O B 2 H 6 + O 2  B 2 O 3 + H 2 O Chemical equation Balanced chemical equation X g O 2 = 10 kg B 2 H 6 (1000 g B 2 H 6 ) (1 mol B 2 H 6 ) (3 mol O 2 ) (32 g O 2 ) (1 kg B 2 H 6 ) (28 g B 2 H 6 ) (1 mol B 2 H 6 ) (1 mol O 2 ) X = 34,286 g O2O2 10 kg X g

139. 138 A solution is prepared by dissolving 3.73 grams of AlCl 3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. What type of problem(s) is this? Molarity followed by dilution. Solutions

140. 139 A solution is prepared by dissolving 3.73 grams of AlCl 3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. 1 st : = mol L 3.73 g 200.0 x 10 - 3 L 0.140 2 nd : M 1 V 1 = M 2 V 2 (0.140 M)(10.0 mL) = (? M)(100.0 mL) 0.0140 M = M 2 molar mass of AlCl 3 dilution formula final concentration Solutions

141. 140 50.0 mL of 6.0 M H 2 SO 4 (battery acid) were spilled and solid NaHCO 3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO 3 must be used? H 2 SO 4 (aq) + 2NaHCO 3  2H 2 O(l) + Na 2 SO 4 (aq) + 2CO 2 (g) Solution Stoichiometry

142. 141 50.0 mL 6.0 M ? g Look! A conversion factor! 50.0 mL of 6.0 M H 2 SO 4 (battery acid) were spilled and solid NaHCO 3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO 3 must be used? H 2 SO 4 (aq) + 2NaHCO 3  2H 2 O(l) + Na 2 SO 4 (aq) + 2CO 2 (g) Solution Stoichiometry = Our Goal

143. 142 50.0 mL 6.0 M ? g 50.0 mL of 6.0 M H 2 SO 4 (battery acid) were spilled and solid NaHCO 3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO 3 must be used? H 2 SO 4 (aq) + 2NaHCO 3  2H 2 O(l) + Na 2 SO 4 (aq) + 2CO 2 (g) Solution Stoichiometry = Our Goal = g NaHCO 3 H 2 SO 4 50.0 mL 1 mol H 2 SO 4 NaHCO 3 2 mol NaHCO 3 84.0 g mol NaHCO 3 50.4

144. 143 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. First write a balanced Equation. ____NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1

145. 144 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. Now, let’s get organized. Place numerical Information and accompanying UNITS below each compound. ____NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1 0.102 M ? mL 35.0 mL Since 1 L = 1000 mL, we can use this to save on the number of conversions Our Goal

146. 145 Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. Now let’s get to work converting. ____NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1 0.102 M ? mL 35.0 mL = mL NaOH H 2 SO 4 35.0 mL H 2 SO 4 0.125 mol 1000 mL H 2 SO 4 NaOH 2 mol 1 mol H 2 SO 4 1000 mL NaOH 0.102 mol NaOH 85.8 Units Match Solution Stoichiometry: shortcut