Slide 1 / 31 New Jersey Center for Teaching and Learning Progressive Science Initiative This material is made freely available at www.njctl.org and is intended for the non-commercial use of students and teachers. These materials may not be used for any commercial purpose without the written permission of the owners. NJCTL maintains its website for the convenience of teachers who wish to make their work available to other teachers, participate in a virtual professional learning community, and/or provide access to course materials to parents, students and others. Click to go to website: www.njctl.org Slide 2 / 31 AP Chemistry Unit 3: Presentation B Chemical Reactions and Stoichiometry www.njctl.org Slide 3 / 31 Chemical Reactions and Stoichiometry The world is full of chemical reactions. They power our batteries, allow us to see, and make nice red paint for sports cars! light cis-retinal trans-retinal The conversion of cis-retinal to trans-retinal is the first step of many that allows us to see.
Slide 4 / 31 Chemical Reactions To review, all chemical reactions must obey the law of conservation of mass - they must be "balanced". 3H 2 (g) + N 2 (g) --> 2NH 3 (g) Reactants Products 6 H atoms 6 H atoms 2 N atoms 2 N atoms This supports the Bohr notion that atoms are simply rearranged in a chemical reaction, not created or destroyed. Nuclear reactions do "destroy" atoms as we shall see but still obey the law of conservation of mass Slide 5 / 31 Chemical Reactions When balancing a reaction, only the coefficients may be changed as changing the subscripts changes the nature of the material itself. H 2 (g) + O 2 (g) --> H 2 O(g) One cannot balance the O atoms by making O 2 suddenly O! Singular O is not reacting here, O 2 gas is! Instead, one must change the number of oxygen or water molecules that react. H 2 (g) + 1/2 O 2 (g) --> H 2 O(g) Slide 6 / 31 Chemical Reactions Make sure to write the correct formulas of reactants and products NH 4 CO 3 --> NH 3 + CO 2 + H 2 O Incorrect formula makes balancing impossible (NH 4 ) 2 CO 3 --> 2NH 3 + CO 2 + H 2 O Correct formula makes things easy!!
Slide 7 / 31 Chemical Reactions Fractions may be used as coefficients which can then be multiplied through by a number to get whole number coefficients. 2 NaCl(s) + H 2 O(g) + SO 2 (g) + 1/2 O 2 (g) --> 2 HCl(g) + Na 2 SO 4 To get whole number coefficients, multiply all coefficients by 2! 4 NaCl(s) + 2 H 2 O(g) + 2 SO 2 (g) + O 2 (g) --> 4 HCl(g) + 2 Na 2 SO 4 Slide 8 / 31 Reaction Stoichiometry Coefficients represent the relative number of molecules, elements, or compounds involved in the reaction. 4 NaCl(s) + 2 H 2 O(g) + 2 SO 2 (g) + O 2 (g) --> 4 HCl(g) + 2 Na 2 SO 4 4 moles of NaCl(s) produce 2 moles of Na 2 SO 4 0.1 moles of NaCl(s) produce ______ moles Na 2 SO 4 4.5 x 10 23 formula units NaCl(s) produce ___________ formula units Na 2 SO 4 *Note, the coefficients DO NOT represent the mass ratios - ie. 4 grams of NaCl will NOT produce 4 grams of HCl. Slide 9 / 31 Reaction Stoichiometry Coefficients can be used to determine the relative amounts of substances involved in a reaction. 4 NaCl(s) + 2 H 2 O(g) + 2 SO 2 (g) + O 2 (g) --> 4 HCl(g) + 2 Na 2 SO 4 How many grams of oxygen would be needed to produce 3 moles of HCl(g)? 3 moles of HCl x 1 mol O 2 x 32 g O 2 = 24 g O 2 4 mol HCl 1 mol O 2 molar ratio from balanced equation
Slide 10 / 31 1 What is the proper coefficient in front of the oxygen molecule after the following equation is balanced? NH 3 (g) + O 2 (g) --> NO(g) + H 2 O(l) Slide 11 / 31 2 The fermentation of sugar produces ethyl alcohol and carbon dioxide gas. How many moles of carbon dioxide would be produced from the fermentation of 0.4 mol of glucose? C 6 H 12 O 6 (s) --> C 2 H 5 OH + CO 2 Slide 12 / 31 3 What is the sum total of coefficients when the equation below is balanced and all coefficients are simplified to the lowest whole number ratio? C 2 H 5 NH 2 + O 2 --> CO 2 + N 2 + H 2 O
Slide 13 / 31 4 Given the UNBALANCED reaction below, determine how many L of nitrogen gas would be produced when 110 grams of NaN 3 decompose completely @STP? NaN 3 (s) ---> Na(s) + N 2 (g) A 22.4 L B 44.8 L C 33.1 L D 57.2 L E 11.2 L Slide 14 / 31 5 Phosphorus pentachloride decomposes into phosphorus trichloride gas and chlorine gas. How many total moles of gas will be produced after 50% of a 414 gram sample of phosphorus pentachloride decomposes? A 0.5 moles B 1.0 moles C 1.5 moles D 2.0 moles E 3.0 moles Slide 15 / 31 6 If a 20 mL solution of 0.3 M Ca(NO 3 ) 2 were mixed with a solution of 3 M NaOH, how many mL of the 3 M NaOH solution would be needed to react with all of the calcium ions? Ca 2+ (aq) + 2OH - (aq) --> Ca(OH) 2 (s)
Slide 16 / 31 Reaction Stoichiometry The theoretical yield is the anticipated amount of product that should be made based on the amounts of reactants used and reaction conditions. CaCO 3 (s) --> CO 2 (g) + CaO(s) What is the theoretical yield of CaO (in moles) if 200 grams of calcium carbonate completely decompose? 200 g CaCO 3 x 1 mol CaCO 3 x 1 mol CaO = 2 mol CaO 100 g CaCO 3 1 mol CaCO 3 Slide 17 / 31 Reaction Stoichiometry When two or more reactants are present, the reactant that is used up first (the limiting reactant) will determine the theoretical yield of product. N 2 (g) + 3H 2 (g) --> 2NH 3 (g) If 44.8 L of nitrogen gas reacts with 44.8 L of hydrogen gas @STP, what is the theoretical yield of ammonia? This type of problem can be easily solved via a series of steps. Slide 18 / 31 Reaction Stoichiometry Steps STEP 1: Convert known quantities to moles N 2 (g) + 3H 2 (g) --> 2NH 3 (g) If 44.8 L of nitrogen gas reacts with 44.8 L of hydrogen gas @STP, what is the theoretical yield of ammonia? 44.8 L N 2 (g) = 2 mol N 2 (g) 44.8 L H 2 (g) = 2 mol H 2 (g) One might think that since both reactants are found in the same quantities, we would run out of both at the same time. Why is this not true for this reaction?
Slide 19 / 31 Reaction Stoichiometry Steps STEP 2: Choose a reactant and determine how much of the other would be needed and compare to how much of that reactant is available. N 2 (g) + 3H 2 (g) --> 2NH 3 (g) If 44.8 L of nitrogen gas reacts with 44.8 L of hydrogen gas @STP, what is the theoretical yield of ammonia? Moles Moles of other reactant Reactant available needed N 2 (g) 2 moles 2 mol N 2 x 3/1 = 6 mol H 2 2 mol H 2 x 1/3 = 0.66 mol H 2 (g) 2 moles N 2 Since 6 mol of H 2 are required and only 2 moles are available, H 2 is the limiting reactant. Slide 20 / 31 Reaction Stoichiometry Steps STEP 3: Use limiting reactant amount to determine the theoretical yield of product or amount of excess reactant that reacted. N 2 (g) + 3H 2 (g) --> 2NH 3 (g) If 44.8 L of nitrogen gas reacts with 44.8 L of hydrogen gas @STP, what is the theoretical yield of ammonia? Finding Theoretical Yield 2 mol H 2 x 2 mol NH 3 = 1.33 mol NH 3 3 mol H 2 Finding Amount of Excess Reactant that Reacted 2 mol H 2 x 1 mol N 2 = 0.66 mol N 2 reacted 3 mol H 2 Slide 21 / 31 7 How many grams of excess reactant will remain if 4 grams of hydrogen gas react with 16 grams of oxygen gas by the reaction below? 2H 2 (g) + O 2 (g) --> 2H 2 O(g)
Slide 22 / 31 8 If 16 grams of methane react with 32 grams of oxygen gas by the reaction below, what would be the theoretical yield of carbon dioxide ( in grams)? CH 4 (g) + 2O 2 (g) --> CO 2 (g) + 2H 2 O(g) A 2 grams B 11 grams C 22 grams D 44 grams E 0.5 grams Slide 23 / 31 9 How many total liters of gas can be produced @STP if 4.8 grams of magnesium metal react with 100 mL of 0.5 M HCl? Mg(s) + 2H + (aq) --> Mg 2+ (aq) + H 2 (g) Slide 24 / 31 10 If 8 molecules of propane (C 3 H 8 ) react with 10 molecules of oxygen gas, which would be true of the reaction mixture after the reaction had gone to completion? C 3 H 8 (g) + 5O 2 (g) --> 3CO 2 (g) + 4H 2 O(g) C 3 H 8 O 2 CO 2 H 2 O Answer A 0 3 2 2 B 6 0 6 8 2 5 3 4 C
Slide 25 / 31 11 If 300 mL of 0.2 M Ca(NO 3 ) 2 solution is mixed with 200 mL of 0.3 M Na 3 PO 4 solution, what will be the concentration of phosphate ions after the reaction below has gone to completion? 3Ca 2+ (aq) + 2PO 43- (aq) --> Ca 3 (PO 4 ) 2 (aq) Slide 26 / 31 Reaction Stoichiometry The amount of product made in the laboratory is often less than what is theoretically possible. Actual Yield x 100 = % Yield Theoretical Yield When 30 grams of CaCO 3 (s) decompose by the following reaction and produce 10 grams of CaO, what is the % yield? CaCO 3 (s) --> CO 2 (g) + CaO(s) 30 g CaCO 3 x 1 mol x 1 mol CaO x 56 g = 16.8 g CaO 100g CaCO 3 1 mol CaCO 3 1 mol CaO 10 g CaO Actual Yield x 100 = 59.5% yield 16.8 g CaO Theoretical Yield Slide 27 / 31 Reaction Stoichiometry If the yield of a reaction is known, it can be used to modify the amounts of reactants needed to make the required amount of product. How many grams of NaN 3 (s) must be decomposed to produce 62 L of N 2 (g) to fill an airbag at a 89% yield @STP? 2NaN 3 (s) --> 2Na(s) + 3N 2 (g) 62 L Actual Yield N2 x 100 L Theoretical Yield N 2 = 69.7 L 89 L Actual Yield N 2 We must plan to make 69.7 L instead of 62 L N 2 . 69.7 L N 2 x 1 mol N 2 x 2 mol NaN 3 x 65 g NaN 3 = 135 g NaN 3 22.4 L N 2 3 mol N 2 1 mol NaN 3
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