Limiting Reagent

LIMITING REAGENT PROBLEMS The first step in solving a limiting reagent problem is being able to recognize that you have a limiting reagent problem. Suppose you were given the following problem: A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl according to the reaction: Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O What is the theoretical yield of MgCl2? Is this a limiting reagent problem? One way to find out is to write down what is known about any component of the reaction below that component: 26670007810500Mg(OH)2 + 2 HCl MgCl2 + 2 H2O 50.6 g 45.0 g ? g Notice how quantities of both reactants are known. Which one will be used up first? You can't tell, nor should you jump to any conclusions. Just because it looks like there is less Mg(OH)2 present does not automatically mean it will be used up before all of the HCl is consumed. This is a limiting reagent problem. There are numerous ways to solve a limiting reagent problem. I will present 3 different methods. You choose the method that you feel most comfortable with. Method 1 In order to find out which reactant is the limiting reagent, you have to compare them to each other. This comparison must be done in moles; therefore, the next step will be to convert each of the grams of reactants to moles using their molar mass: Again, you should not jump any conclusions about which reactant is the limiting reagent. Just because there are fewer moles of magnesium hydroxide does not mean it is the limiting reagent. Arbitrarily pick one of these reactants and calculate how many moles of the other reactant is needed to completely use up the reactant picked. In this case, magnesium hydroxide is arbitrarily chosen: Compare the moles HCl needed to the actual moles HCl available. In this case, 1.74 mole of HCl is needed and 1.23 mole HCl is available--that's not enough. So, even though it appears that there are more moles of HCl than Mg(OH)2, the HCl is the limiting reagent. The HCl will be run out before the magnesium hydroxide and thereby limit the amount of product formed. For this reason, use the moles of HCl to calculate the theoretical yield of magnesium chloride: The theoretical yield is the maximum amount of product that can be produced (in an ideal world). In the "real" world it is difficult to produce the amount obtained for the theoretical yield. A percent yield is often used to show how close to ideality one has obtained in a chemical synthesis. Suppose in the reaction discussed a chemist actually obtained 55.4 g of MgCl2. This is called the actual yield and would be given to you in the problem. To calculate the percent yield: Method 2 An alternative method involves comparing the theoretical ratio of the reactants to the actual ratio of reactants available. First, calculate the moles of each reactant using their molar mass: available available Consider the balanced reaction: Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O From the balanced equation, the theoretical mole ratio is: Let’s see what the actual mole ratio is based on actual the amounts of reactants present: Comparing these two ratios, one can easily see that there is not enough HCl, so HCl must be the limiting reagent: Since HCl is the limiting reagent, use the moles of HCl available to calculate the theoretical yield of MgCl2: Method 3 A third method is to calculate the theoretical yield of product produced by each reactant and chose the lesser amount: Since HCl produced less product, HCl is the limiting reagent and 58.6 g MgCl2 is the theoretical yield. Stoichiometry Problem Set 1. Calculate the molar mass of the following compounds to two decimal places. a. AlCl3 b. Na2SO4 c. Cr2(CO3)3 d. C7H5N3O6 e. CuSO4?5H2O 2. Given 115 grams of Na2SO4, find: a. the number of moles of sodium sulfate. b. the number of moles of sodium ions c. the number of moles of sulfate ions d. the number of moles of sulfur e. the number of moles of oxygen 3. Given 3.5 moles of K2CO3, find: a. the number of grams of potassium carbonate b. the number of moles of potassium ions c. the number of moles of carbonate ions d. the number of potassium ions e. the number of carbon atoms 4. Balance the following reactions by inspection: a. S8 + F2 ? SF4 b. CH4 + Cl2 ? CCl4 + HCl c. C5H12 + O2 ? CO2 + H2O d. Ba(NO3)2 + Na2CO3 ? BaCO3 + NaNO3 e. CaCl2 + K3PO4 ? Ca3(PO4)2 + KCl 5. If 12.5 grams of propane, C3H8, react with excess oxygen according to the following unbalanced reaction: C3H8 + O2 ? CO2 + H2O Calculate the following: a. the moles of oxygen gas needed b. the moles of carbon dioxide formed c. the grams of carbon dioxide formed d. the percent yield, if 35.2 grams of carbon dioxide are actually isolated e. the molecules of carbon dioxide formed 6. If 5.39 x 1024 copper atoms react with excess nitric acid according to the reaction: Cu(s) + 4 HNO3(aq) ? Cu(NO3)2(aq) + 2NO2(g) + 2 H2O(l) Calculate the following: a. the number of grams of copper atoms b. the number of moles of nitric acid needed c. the number of grams of copper(II) nitrate formed d. the number of molecules of nitrogen dioxide formed e. the number of moles of water formed 7. If 6.20 grams of ethane, C2H6, reacts with 92.0 grams bromine liquid to form hexabromoethane, C2Br6 and hydrogen bromide gas according to the following reaction: C2H6 (g) + 6 Br2 (l) ? C2Br6 (l) + 6 HBr (g) a. Find the limiting reactant b. Calculate the moles of HBr formed c. Calculate the theoretical yield of C2Br6 d. If the actual yield of hexabromoethane is 45.9 grams, what is the percent yield? e. Calculate the grams of reactant left over in excess 8. If 14.7 grams of copper reacts with 450.0 mL of 2.50 M HNO3 according to the reaction: Cu(s) + 4 HNO3(aq) ? Cu(NO3)2(aq) + 2NO2(g) + 2 H2O(l) Calculate the following: a. the grams of copper(II) nitrate formed b. the molarity of the the copper(II) nitrate in solution (assume volume of Cu is negligible) c. the molecules of nitrogen dioxide formed d. the moles of excess reactant e. If 39.5 grams of copper(II) nitrate is isolated, what is the percent yield? Answers Most stoichiometry problems follow the strategy: Quantity A ? mol A ? mol B? Quantity B You will be using portions of this strategy or the entire strategy to complete this problem set. 1. Sum the atomic weights (AW) of each of the elements in the compound according to the number of each element present. The molar mass is defined as the number of grams of substance per mole of substance. It has the same mass in g/mol as one formula unit in amu. a. AlCl3 Molar mass = AW Al + 3(AW Cl) = 26.98 g/mol + 3(35.45 g/mol) = 133.33 g/mol b. Na2SO4 Molar mass = 2(AW Na) + AW S + 4(AW O) = 2(22.99 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 142.05 g/mol c. Cr2(CO3)3 Molar Mass = 2(AW Cr) + 2(AW C) + 6(AW O) = 2(52.00 g/mol) + 3(12.01 g/mol) + 9(16.00 g/mol) = 284.03 g/mol d. C7H5N3O6 Molar Mass = 7(AW C) + 5(AW H) + 3(AW N) + 6(AW O) = 7(12.01 g/mol) + 5(1.01 g/mol) + 3(14.01 g/mol) + 6(16.00 g/mol) = 227.15 g/mol e. CuSO4?5H2O (a hydrate) Molar Mass = AW Cu + AW S + 9(AW O) + 10(AW H) = 63.54 g/mol + 32.07 g/mol + 10(1.01 g/mol) + 9(16.00 g/mol) = 249.71 g/mol 2. a. Use the molar mass of sodium sulfate as a “conversion factor” to calculate the moles of sodium sulfate. In terms of the stoichiometry strategy, the molar mass allows you to convert a Quantity A to mol A or mol B to Quantity B. There are 142.05 grams of Na2SO4 in 1 mole of Na2SO4. Let the units decide how to set up the problem. 115 g Na2SO4 x1 mol Na2SO4142.05 g Na2SO4 = 0.810 Na2SO4 b. Here we are taking Quantity A to mol A to mol B. The subscript of 2 in the formula for Na2SO4 indicates there are 2 moles of sodium ions per mole of sodium sulfate. Let the units help you set up the problem: 115 g Na2SO4 x1 mol Na2SO4142.05 g Na2SO4 x 2 mol Na+1 mol Na2SO4 = 1.62 mol Na+ c. Again, here we are taking Quantity A to mol A to mol B. In mole of sodium sulfate there is one mole of sulfate ion. The units help you set up the problem in a similar fashion as in problem 2a: 115 g Na2SO4 x1 mol Na2SO4142.05 g Na2SO4 x 1 mol SO42-1 mol Na2SO4 = 0.810 mol SO42- d. In one mole of sodium sulfate there is 1 mole of sulfur as indicated by the subscript of one in the formula. Once again, units help set up the problem for converting Quantity A to mol A to mol B: 115 g Na2SO4 x1 mol Na2SO4142.05 g Na2SO4 x 1 mol S1 mol Na2SO4 = 0.810 mol S e. In mole of sodium sulfate there are four moles of oxygen as indicated by the subscript of four in the formula. Once again, units help set up the problem for converting Quantity A to mol A to mol B: 115 g Na2SO4 x1 mol Na2SO4142.05 g Na2SO4 x 4 mol O1 mol Na2SO4 = 3.24 mol O 3 a. Here we already know mol A and will convert to mol B to Quantity B of the strategy. We will need the molar mass of potassium carbonate to do the mol B to Quantity B conversion: Molar mass = 2(AW K) + AW C + 3(AW O) = 2(39.10 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 138.21 g/mol Let the units help you set up the problem. One mole of potassium carbonate has a mass of 138.21 grams: 3.5 mol K2CO3 x 138.21 g K2CO31 mol K2CO3 = 4.8 x 102 g K2CO3 b. Convert mol A to mol B. In one mole of potassium carbonate there are two potassium ions: 3.5 mol K2CO3 x 2 mol K+1 mol K2CO3 = 7.0 mol K+ c. Once again, convert mol A to mol B. In one mole of potassium carbonate there is one mole of carbonate: 3.5 mol K2CO3 x 1 mol CO32-1 mol K2CO3 = 3.5 mol CO32- d. Here, the number of particles of potassium ions is required. This tips you off that Avogadro’s number is needed somewhere in the problem. Avogadro’s number is the number of particles in one mole of substance. There will be 6.02 x 1023 potassium ions in one mole of potassium ions. Convert mol A to mol B to Quantity B. In this case, mol potassium carbonate to mol potassium ions to number of potassium ions: 3.5 mol K2CO3 x 2 mol K+1 mol K2CO3 x 6.02 x 1023 K+ ions1 mol K+ = 4.2 x 1024 K+ ions e. This problem is similar to 3d. Convert mol A to mol B to Quantity B. In this case, mol potassium carbonate to mol carbon to number of carbon atoms: 3.5 mol K2CO3 x 1 mol C1 mol K2CO3 x 6.02 x 1023 C atoms1 mol C = 2.1 x 1024 C atoms 4. When balancing chemical reactions there are a few things to keep in mind. Never change the subscripts in the formulas found in the reaction. Place coefficients in front of the substance, never in the middle of the substance. Polyatomic ions may be balanced as a unit, if they are unchanged on both sides of the reaction. There are a few tricks one can use to quickly balance chemical reactions. If an element is off by itself, balance that element last. You can always put any number needed in the end to balance that element. If there is an odd number of an element on one side the reaction, and an even number on the other, make the odd side even by multiplying by two. a. Balance the sulfur first, as there is an odd number on the right hand side and an even number on the left hand side. In this case, don’t multiply by 2, rather balance it with an 8: S8 + F2 ? 8 SF4 Now the sulfur is balance, but there are 2 F on the left hand side and 32 F on the right hand side. Balance the F with a coefficient of 16 on the left hand side: S8 + 16 F2 ? 8 SF4 b. Balance the Cl last, as it found as an element on the left hand side. The carbon is already balanced, so balance the H. There are 4 H on the left hand side, so place a coefficient of 4 with the HCl to balance the H: CH4 + Cl2 ? CCl4 + 4 HCl The carbon is still balanced along with the H. There are 2 Cl on the left hand side and 8 Cl on the right hand side Don’t forget to look at all the products – 4 Cl are found in CCl4 and 4 Cl are with the 4 HCl. Balance the Cl by placing a coefficient of 4 with the Cl on the left hand side: CH4 + 4 Cl2 ? CCl4 + 4 HCl c. Oxygen is off by itself, so balance it last. There are 5 C on the left hand side and 1 C on the right hand side. Balance the carbon by placing a coefficient of 5 in front of CO2: C5H12 + O2 ? 5 CO2 + H2O Next balance the H. There are 12 H on the left hand side and 2 H on the right hand side. Place a coefficient of 6 in front of H2O to balance the H: C5H12 + O2 ? 5 CO2 + 6 H2O Lastly, balance the O. There are 2 O on the left hand side and 16 O on the right hand side. Place a coefficient of 8 in front of O2 to balance the O: C5H12 + 8 O2 ? 5 CO2 + 6 H2O d. In this problem, there isn’t any easy place to start. To keep things simpler, notice that the polyatomic ions can be balanced as the ion, rather than as separate elements. Barium is balanced, but there are 2 NO3? ions on the left hand side and 1 NO3? ion on the right hand side. Place a coefficient of 2 in front of NaNO3 to balance nitrate: Ba(NO3)2 + Na2CO3 ? BaCO3 + 2 NaNO3 This also balances the Na. There are now 2 Na on both sides of the reaction. Also, CO32- is balanced as well. There is one carbonate in both sides of the reaction. e. Once again, there isn’t an easy place to start, but you should recognize that phosphate ions, PO43- can be balanced as the polyatomic ion. Start with balancing the calcium (arbitrary). There is 1 Ca on the left hand side and 3 Ca on the right hand side. Place a coefficient of 3 in front of CaCl2 to balance the calcium: 3 CaCl2 + K3PO4 ? Ca3(PO4)2 + KCl From here, continue on sequentially. There are now 6 Cl on the left hand side and 1 Cl on the right hand side. Balance the Cl by placing a coefficient of 6 in front of KCl to balance the Cl: 3 CaCl2 + K3PO4 ? Ca3(PO4)2 + 6 KCl Now there are 6 K on the right hand side and 3 K on the left hand side. Balance the K by placing a coefficient of 2 in front of K3PO4: 3CaCl2 + 2 K3PO4 ? Ca3(PO4)2 + 6 KCl This also balances the PO43-. There are now 2 phosphate ions on the left hand side and 2 phosphate ions on the right hand side. 5. Balance the reaction first: C3H8 + 5 O2 ? 3 CO2 + 4 H2O. We will be back to using the stoichiometry strategy, Quantity A ? mol A ? mol B ? Quantity B. a. The molar mass of propane, C3H8, will be needed: Molar mass = 3(AW C) + 8(AW H) = 3(12.01 g/mol) + 8(1.01 g/mol) = 44.11 g/mol Use the molar mass to convert grams of propane to mol of propane and the balance reaction to convert mol of propane to mol of oxygen gas needed. From the strategy, we are using Quantity A ? mol A ? mol B: 12.5 g C3H8 x 1 mol C3H8 44.11 g C3H8 x 5 mol O21 mol C3H8 = 1.42 mol O2 b. From the balanced reaction, 3 moles of carbon dioxide are formed from 1 mole of propane. Once again, we are converting grams of propane to moles of propane using its molar mass and then converting moles of propane to moles of carbon dioxide. From the strategy, we are using Quantity A ? mol A ? mol B: 12.5 g C3H8 x 1 mol C3H8 44.11 g C3H8 x 3 mol CO21 mol C3H8 = 0.850 mol CO2 c. Here we will use the entire strategy. Convert the grams of propane to moles of propane, then moles of propane to moles of carbon dioxide and finally moles of carbon dioxide to grams of carbon dioxide. We will need to calculate the molar mass of carbon dioxide. All other information has already been previously found in parts a – c of the problem: Molar mass of CO2 = AW C + 2(AW O) = 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol 12.5 g C3H8 x 1 mol C3H8 44.11 g C3H8 x 3 mol CO21 mol C3H8 x 44.01 g CO21 mol CO2 = 37.4 g CO2 d. The mass of carbon dioxide calculated in part c of this problem is often referred to as the theoretical yield. The theoretical yield represents the maximum amount of product that can be produced in an ideal world with ideal chemists who can collect every single particle of product formed. Needless to say in the real world this doesn’t happen. As a result one often can calculate a percent yield. A percent is always a “part/whole” ratio. The actual yield, 35.2 g CO2 represents the “part” and the theoretical yield, 37.4 g CO2, represents the “whole.” To calculate the percent yield: % yield = actual yieldtheoretical yield x 100 = 35.2 g CO237.4 g CO2 x 100 = 94.1 % yield e. Once again, the entire strategy is used. Since we are asked for the number of molecules (particles) of CO2, Avogadro’s number will be needed. Convert the mass of C3H8 to moles of C3H8 to moles of CO2 to molecules of CO2: 12.5 g C3H8 x 1 mol C3H8 44.11 g C3H8 x 3 mol CO21 mol C3H8 x 6.02 x 10231 mol CO2 = 5.1 x 1023 molecules CO2 6. a. Convert the number of copper atoms to moles copper to grams copper: 5.39 x 1024 Cu atoms x 1 mol Cu6.02 x 1023 Cu atoms x 63.54 g Cu1 mol Cu = 569 g Cu b. Convert the number of copper atoms to moles copper to moles of nitric acid: 5.39 x 1024 Cu atoms x 1 mol Cu6.02 x 1023 Cu atoms x 4 mol HNO31 mol Cu = 35.81 mol HNO3 c. Convert the number of copper atoms to moles copper to moles copper(II) nitrate to grams of copper(II) nitrate using the balanced reaction: 5.39 x 1024 Cu atoms x 1 mol Cu6.02 x 1023 Cu atoms x 1 mol Cu(NO3)21 mol Cu x 187.56 g Cu(NO3)21 mol Cu(NO3)2 = 1.68 x 103 g Cu(NO3)2 d. Since molecules are particles and directly analogous to moles, one can convert molecules copper directly to molecules nitrogen dioxide using the coefficients in the balanced reaction: 5.39 x 1024 Cu atoms x 2 NO2 molecules 1 Cu atom = 1.08 x 1025 NO2 molecules e. Convert the atoms of copper to moles of copper to moles of water using the balanced reaction: 5.39 x 1024 Cu atoms x 1 mol Cu6.02 x 1023 Cu atoms x 2 mol H2O1 mol Cu = 17.9 mol H2O 7. A limiting reactant problem is recognizable because quantities of both reactants are given in the problem. One of these reactants will be used up before the other is totally consumed. a. To find the limiting reactant, first find the moles of each reactant so that they may be compared. Don’t jump to any conclusions about the limiting reactant. The reactant with a lower mass isn’t always the limiting reactant. 6.20 g C2H6 x 1 mol C2H630.08 g C2H6 = 0.206 mol C2H6 92.0 g Br2 x 1 mol Br2159.82 g Br2 = 0.576 mol Br2 Once again, don’t jump to any conclusion about the limiting reactant. Even though it appears that there are fewer moles of ethane, a comparison has not been made to establish which reactant will be consumed first. Pick one of the reactants and calculate how many moles of the other reactant is needed. Let’s arbitrarily choose ethane and calculate the moles of bromine needed using the balanced reaction: 0.206 mol C2H6 x 6 mol Br2 needed1 mol C2H6 = 1.24 mol Br2 needed Although it appears that there is a greater amount of bromine present, bromine turns out to be the limiting reactant: 0.618 moles of bromine are needed to react all 0.206 mol of ethane, and only 0.576 mole of bromine is available. b. Use the moles of limiting reactant and the balanced reaction to calculate the moles of HBr formed: 0.576 mol Br2 x 6 mol HBr6 mol Br2 = 0.576 mol HBr c. Using the moles of limiting reactant and the balanced reaction to calculate the theoretical yield in grams of C2Br6: 0.576 mol Br2 x 1 mol C2Br66 mol Br2 x 503.74 g C2Br61 mol C2Br6 = 48.4 g C2Br6 d. A percent is always a “part/whole” ratio. Use the actual yield as the “part” and the theoretical yield as the “whole”: % yield = actual yieldtheoretical yield x 100 = 45.9 g actual96.7 g theoretical x 100 = 94.9 % yield e. The reactant in excess was ethane. Find the moles of ethane that actually reacted using the balanced reaction: 0.576 mol Br2x 1 mol C2H6 used 6 mol Br2 = 0.0960 mol C2H6 used Subtract the mole ethane used from the total moles of ethane available: 0.206 mol C2H6 – 0.0960 mol C2H6 = 0.110 mol C2H6 in excess Convert the moles of ethane in excess to grams: 0.110 mol C2H6 x 30.08 g C2H61 mol C2H6 = 3.31g C2H6 in excess 8. a. This is a limiting reactant problem because the quantities of both reactants are given. It is impossible to tell which reactant is used up before the other. A comparison of both reactants must be made in moles to determine the limiting reactant. Convert each of the reactant quantities to moles. The atomic weight of copper is used to convert grams copper to moles copper: 14.7 g Cu x 1 mol Cu63.54 g Cu = 0.231 mol Cu Think of molarity as a conversion factor that allows you to convert moles of solute to liters of solution or in this case, volume of solution to moles of solute: 450.0 mL soln x 1 x 10-3 L1 mL soln x 2.50 mol HNO31 L soln = 1.12 mol HNO3 Choose one reactant and calculate the moles of the other reactant needed. In this case, mole of copper is arbitrarily chosen and the mole of nitric acid needed is calculated: 0.231 mol Cu x 4 mol HNO3 needed 1 mol Cu = 0.924 mol HNO3 needed Compare the moles of nitric acid needed to the moles of nitric acid available: 0.924 mol of HNO3 are needed and 1.12 mol of HNO3 is available. It is clear that HNO3 is in excess, thus Cu is the limiting reactant. Use the mole of limiting reactant to calculate the theoretical yield in grams of copper(II) nitrate formed: 0.231 mol Cu x 1 mol Cu(NO3)21 mol Cu x 187.56 g Cu(NO3)21 mol Cu(NO3)2 = 43.3 g Cu(NO3)2 b. The total volume of the solution is 450.0 mL since we are assuming the volume copper is not significantly adding to the total volume of the solution. First find the moles of Cu(NO3)2 formed and divide the moles of solute by the volume in liters: 0.231 mol Cu x 1 mol Cu(NO3)21 mol Cu = 0.231 mol Cu(NO3)2 450.0 mL soln x 1 x 10-3 L1 mL soln = 0.4500 L solution 0.231 mol Cu(NO3)20.4500 L soln = 0.531 M Cu(NO3)2 c. Convert the mole of limiting reactant to mole of nitrogen dioxide to molecules of nitrogen dioxoide: 0.231 mol Cu x 2 mol NO21 mol Cu 6.02 x 1)23 molecules NO21 mol NO2 = 2.78 x 1O23molecules NO2 d. The mole of HNO3 needed and the total mole of HNO3 available has already been found in part a of the problem. To find the mole of HNO3 in excess, subtract the mole of HNO3 needed from the total mole of HNO3 available: 1.12 mol HNO3 – 0.924 mol HNO3 = 0.20 mol HNO3 in excess e. A percent is always a “part/whole” ratio. Use the actual yield as the “part” and the theoretical yield as the “whole”: % yield = actual yieldtheoretical yield x 100 = 39.5 g actual43.3 g theoretical x 100 = 91.2 % yield