Mole Stoichiometry Teacher Key

Not all things in life occur exactly as planned. Even when baking a cake, it is highly probable that, (although I consider myself a pretty adept baker), when I measure my 2 cups of butter and 3 cups of flour, I may inadvertently use a little less or a little more than the required amounts of each ingredient. It may be due to the inaccuracy of the measuring tools I am using, or my own human error in reading these measuring tools. We can sometimes see the consequences of this in our final baked dessert. In Chemistry, when we don’t have exactly the amount of each reactant that was calculated in our mole stoichiometry, we will see a different outcome of our reaction. But first...consider the following analogies: MAKING a GRILLED CHEESE SANDWICH: The recipe is: 2 slices of bread 1 slice of cheese 1 sandwich à + But what if I want to open my own food truck? And I want it to be Toombsy’s Cheesy Sandwich Truck. (The name is still being developed Obviously I will have to make more than one sandwich at a time. So, for the first day I go out and buy: 20 slices of bread AND 10 slices of cheese How many sandwiches can I make? 10 The first day is SUCH a success that the next day I have a bulk order delivered and I get: 350 slices of bread AND 200 slices of cheese How many sandwiches can I make? 175 What will be leftover at the end of the day: 25 slices of cheese. ) Don’t like grilled cheese sandwiches? (Seriously?) Then how about constructing bicycles: à + Consider that I have 21 frames 20 frames 1 frame will be leftover We call this the EXCESS REAGENT Consider that I have 40 bicycle tires + 40 bicycle tires All 40 tires will be used How many Bikes will I make? à 20 bikes We call this the LIMITING REAGENT Now let’s try this with MOLE STOICHIOMETRY: 1.00 moles of aluminum phosphate (aq) are reacted with 4.00 moles of solid lithium metal. How many moles of products are formed? AlPO4 (aq) + 3Li (s)à Li3PO4 (aq) + Al (s) 1.00 mol 4.00 mol SCENARIO #1 If the 1.00 mole of AlPO4 (aq)completely reacts, then only 3.00 moles of Li (s) is needed: AlPO4 (aq) + 3Li (s)à Li3PO4 (aq) + Al (s) 1.00 mol 3.00 mol SCENARIO #2 If the 4.00 mole of Li (s) completely reacts, then 1.33 moles of AlPO4 (aq) is needed: AlPO4 (aq) + 3Li (s)à Li3PO4 (aq) + Al (s) 1.33 mol 4.00 mol But clearly, only SCENARIO #1 can happen. SCENARIO #2 is impossible, because not enough AlPO4 (aq) was provided to carry that out. AlPO4 (aq) Is called the LIMITING REAGENT: it limits us to the smaller ratio in SCENARIO #1 The limiting reagent will be completely used up (totally reacted). Li (s) Is called the EXCESS REAGENT: We have 1.00 more mole than we need for SCENARIO #2 The excess reagent will be leftover after the reaction is over (you will see unreacted Li (s) still in the test tube). Limiting and Excess MOLE STOICHIOMETRY: 1) 39.9 g of aluminum chloride reacts with 52.3 grams of sodium carbonate. Step one. Write the balanced chemical reaction: 2AlCl3 + 3Na2CO3 à Al2(CO3)3 + 6NaCl Step two. Convert all given data in moles. 2AlCl3 + 3Na2CO3 à Al2(CO3)3 + 6NaCl 39.9 g X 1 mole 133.341 g =0.300moles This is the amount of AlCl3 that we HAVE 52.3 g X 1 mole 105.988 g =0.493 moles This is the amount of Na2CO3 that we HAVE Step three. Determine limiting and excess reactants (also called reagants) by using mole stoichiometry (ratios) 2AlCl3 + 3Na2CO3 à Al2(CO3)3 + 6NaCl 0.300 moles 0.300 X 3 2 = 0.450 moles needed 0.493 X 2 3 = 0.329 moles needed 0.493 moles NOT POSSIBLE! I don’t have 0.329 moles of AlCl3 so I am LIMITED by the amount of AlCl3 that I have, and the second line can not occur. LIMITING REAGENT IS: AlCl3 EXCESS REAGENT IS: Na2CO3 2AlCl3 0.300 moles + 3Na2CO3 0.450 moles à Al2(CO3)3 + 0.300 X ½ 0.150 moles X 233.988 g/mol = 35.1 g LAW OF CONSERVATION OF MASS : CHECK LS: 39.9 + 52.3 g = 92.2 g RS: = 35.1 + 52.6 = 87.7 g Is there 4.5 g of excess Na2CO3? Let’s check! 6NaCl 0.300 X 3 0.900 mol 58.443 g/mol = 52.6 g 0.493 moles of Na2CO3 available -0.450 moles used up = 0.043 moles of excess unreacted Na2CO3 after the reaction is complete 0.043 moles of excess X 105.988 g/mol = 4.557 g excess, (4.6 g to 2 significant figures) THIS CHECKS OUT! 2) 4.763 g of Titanium is reacted with 13.98 g of copper (II) chloride to form titanium (IV) chloride and copper: Ti + 2CuCl2 à TiCl4 + 2 Cu 4.763 g X 1 mole 47.867 g =0.09950 moles This is what we HAVE 13.98 g X 1 mole 134.452 g =0.104 moles THIS IS WHAT WE HAVE Ti + 2CuCl2 à HAVE 0.09950 moles TiCl4 + 2 Cu need 2 X 0.0995 = 0.199 moles But we don’t have this much We only HAVE 0.104 moles… SO…We would need 0.104 moles X 1 2 = 0.0520 moles A better way to represent this: Ti + 2CuCl2 0.09950 moles Excess reactant is Ti 0.199 moles 0.0520 moles 0.104 moles Limiting reactant is CuCl2 TiCl4 + 2 Cu 0.0520 moles 0.104 moles à Therefore, we HAVE 0.09950 moles of Ti but we will only use up 0.0520 moles There will be an excess of 0.09950 – 0.0520 moles = 0.0475 moles Ti that is unreacted. All 0.104 moles of the CuCl2 will react. This is the limiting reactant. There will be none left as all of it turns into products. We use this value to determine the moles of each product formed. 0.0520 moles TiCl4 x 189.679g TiCl4 = 9.863 g 0.104 moles Cu x 63.546 g Cu = 6.609 g Mole mole LAW OF CONSERVATION OF MASS : CHECK LS: 4.763 g + 13.98 g = 18.743 g RS: = 9.863 g + 6.609 g = 16.472 g BUT 0.0475 moles excess Ti X 47.867 g / mole = 2.274 g excess 18.743 g - 2.274 g excess unreacted Ti = 16.469 g J 33.7 g of aluminum reacts with 114 g of chromium (II) oxide in a single replacement reaction. a) Name the limiting reactant. b) Name the excess reactant. c) Determine the grams of each product formed. d) Show your check of the Law of Conservation of Mass. e) Name the amount of excess reactant that remains, in grams. 2Al We have 33.7 g X 1 mole 26.982g =1.25 moles X 2 Al 3 CrO = 1.12 moles needed + 3 Cr BUT…We have 114 g =1.67 moles X 1 mole 67.995g =1.68 moles 3CrO 1.25 moles excess 1.875 moles 1.12 moles 0.13 moles of excess Al à Al2O3 + 3 Cr We don’t have enough 1.68 moles Limiting ALL 1.68 moles of CrO reacts 30.2 g Al that actually reacts TOTAL: Al2O3 + X 3 CrO 2 Al =1.875 moles needed 2Al + + à 3CrO 114 g CrO 144.2 g 0.560 moles 1.68 moles 0.560 moles Al2O3 x 101.961g = 57.1 g mole 1.68 moles Cr x 51.996 g Cr= 87.4 g mole 144.5 g 1) 111 g of calcium chloride reacts with 100.0 g of sodium phosphide. a) Which reactant is in excess? b) Calculate the mass of excess that remains. c) Calculate the mass of each product that is produced. d) Show your check of the Law of Conservation of Mass. 111 g of CaCl2 X 110.984 g / mole = 1.00 moles 100.0 g of Na3P X 99.9444 g / mole = 1.00 moles 3CaCl2 + 2Na3P 1.00 moles limiting X 3/2 =1.50 moles X 2/3 =0.667 moles à Ca3P2 + 6NaCl 1.00 moles X 1/3 =0.333 moles 1.00 moles X 2 =2.00 moles X 182.182 g / mol = 60.7 g X 58.443 g / mol = 117 g 1.00 moles excess We don’t have enough 1.00 mole available = 100.0 g Only 0.667 moles reacted = 66.7 g 111 g + 66.7 g = 177.7 g (178 g) 2) 60.7 g + 117 g = 177.7 g (178 g) 10.0 g of calcium carbonate reacts with 6.00 g of sodium chloride. a) Which reactant is in excess? b) Calculate the mass of excess that remains. c) Calculate the mass of each product that is produced. d) Show your check of the Law of Conservation of Mass. 10.0 g of CaCO3 X 100.086 g / mole = 0.0999 moles 6.00 g of NaCl X 58.443 g / mole = 0.103 moles CaCO3 + 2NaCl 0.0999 moles excess X2 =0.1998 moles à Na2CO3 + CaCl2 We don’t have enough X½ = 0.0515 moles 0.103 moles limiting 0.515 moles = 0.515 moles X 105.961 g / mol X 110.984 g / mol = 5.46 g = 5.72 g 0.0515 moles X 100.086 g / mole = 5.13 g of CaCO3 reacted 5.13 6.00 g = 11.13g 5.46 g + 5.72 g = 11.18 g 3) A students spills 35.0 g of calcium hydroxide on the lab table and in an effort to neutralize it, they pour 35.0 g of phosphoric acid onto the spill. a) Which reactant is in excess? b) Calculate the mass of excess that remains. c) Calculate the mass of each product that is produced. d) Show your check of the Law of Conservation of Mass. 35.0 g of Ca(OH)2 X 74.092 g / mole = 0.472 moles 35.0 g of H3PO4 X 97.994 g / mole = 0.357 moles 3Ca(OH)2 + 2H3PO4 0.472 moles limiting X 3/2 = 0.536 moles X 2/3 =0.315 moles 0.357 moles excess à Ca3(PO4)2 + 6HOH X 1/3 =0.158 moles X2 =0.945 moles We don’t have enough 0.357 moles of H3PO4 available - 0.315 moles actually used = 0.042 moles of acid leftover, unreacted X 97.994 g / mole = 4.1 g of excess acid 35.0 g + 35.0 g – 4.1 g = 65.9 g X 310.174 g / mol = 49.0 g X 18.015 g / mol = 17.0 g 49.0 g + 17.0 g = 66.0 g 4) 314 g of sulphuric acid reacts with 70.5 g of lithium hydroxide a) Which reactant is in excess? b) What mass of sulphuric acid was used in the reaction? c) What mass of each product is produced? 1) H2SO4 given 314 g + a) Decide on the limiting and excess reagent =2.94 mol 3.20 mol is what we have 6.40 mol is how much we need to react with the 3.20 mol of H2SO4 2.94 mol is the amount we have b) Calculate grams of acid used 1.47 mol reacted X Mole ratio =144 g reacted Only 1.47 mol reacted + 2H2O X =3.20 mol 1.47 mol is how much we need to react with the 2.94 mol of LiOH This is the EXCESS reagent because we have 3.20 mol available, but we are only going to use 1.47 mol of it à Li2SO4 70.5 g Change to moles X Comparing the Mole ratio 2LiOH à ß This is the limiting reagent because we only have 2.94 mol of LiOH, not enough to react with the given amount of acid All of the 2.94 mol reacted 1.47 mol produced c) Calculate X mass of each =162 g product Check: reactants: 144g + 70.5g = 214.5 g products: 162g + 53.0g = 215 g Alternate check ß Can you figure out what I have done here: reactants: 314g + 70.5g = 384.5g products: 162g + 53.0g = 215g + 170g excess acid remaining = 385g 2.94 mol produced X =53.0 g 5) Francium reacts explosively with water. If 325 g of Francium react with 84.5 g of water, will both reactants be totally consumed? What mass of products can we expect from the reaction? Please note that these answers were calculated using a periodic table that had less significant figures. But you should still be able to compare your answers to ensure you were following the correct method. Moles Fr=(1mole/223g)(325g)=1.46 moles Fr Moles H2O = (1 mole / 18.0 g) (84.5 g) = 4.69 moles H2O This line! 2Fr + 2H2O à 2FrOH + 1.46 moles1.46 moles1.46 moles0.730 moles 4.69 moles4.69 moles4.69 moles2.35 moles H2 Water is present in excess; the excess is (4.69 – 1.46) moles = 3.23 moles Mass excess water = (18.0 g / mole) (3.23 moles) = 58.1 g H2O in excess Mass FrOH = (240. g / mole) (1.46 moles) = 350. g FrOH Mass H2 = (2.02 g / mole) (0.730 moles) = 1.48 g H2 Check: Σ mass of reactants = Σ mass of products + excess 325g+84.5g=350.g+1.48g+58.1g 410.g=410.g Good!