Unit 4 Lecture 4 - Limiting Reactants Quantitative Information from Balanced Chemical Equations

Quantitative Information from Balanced Equations Getting information from a balanced chemical equation is a matter of putting together your knowledge of chemical formulas and mass/moles/numbers relationships, and then applying them to a balanced chemical reaction. Example: Propane is a common fuel. How many moles of oxygen are needed to burn 3.62 moles of propane? 1. Write and balance the equation for the reaction: 3.62 mol ? mol C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l) 2. Write what you know. Write what you need to know.

Quantitative Information from Balanced Equations 3.62 mol ? mol C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l) 3. Convert your “knowns” to moles, if necessary. 4. Use the appropriate stoichiometric equivalences from the equation to convert from moles of what you have to moles of what you need. 5. Convert your answer from moles to the required units. 3.62 mol C3H8 x 5 mol O2 = 18.1 mol O2 1 mol C3H8 stoichiometric equivalence (from chemical equation) Here we were asked for moles, so no mole/mass conversion was necessary.

Quantitative Information from Balanced Equations Example: What mass of oxygen is consumed in the combustion of 100.0 g of propane? 1. Write and balance the equation for the reaction. 100.0 g ? g C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l) 2. Write what you know and what you need to know. 3. Convert “what you know” to moles. 4. Using the stoichiometry of the reaction, convert from moles of “what you know” to moles of “what you need to know.” 5. Convert the moles of what you need to appropriate final units. 100.0 g C3H8 x 1 mol C3H8 x 5 mol O2 x 32.00 g O2 = 362.8 g O2 44.10 g 1 mol C3H8 1 mol O2 mass  moles stoichiometric equivalence moles  mass

Unit 4 Lecture 4 - Limiting Reactants Quantitative Information from Balanced Equations - Example Detonation of nitroglycerin proceeds as follows: 4C3H5N3O9(l)  12CO2(g) + 6N2(g) + O2(g) + 10H2O(g) a) If a sample containing 2.00 mL of nitroglycerin (density = 1.592 g/mL) is detonated, how many total moles of gas are produced? b) If each mole of gas occupies 55 L under the conditions of the explosion, how many liters of gas are produced? c) How many grams of N2 are produced in the detonation? answers: a) 0.102 mol, b) 5.6 L, c) 0.589 g

Limiting Reactant N2 (g) + 3H2 (g)  2NH3 (g) If we start with 2 moles nitrogen and 6 moles hydrogen, how many moles of ammonia is it theoretically possible to produce? One way to work the problem: 2 mol N2 x 2 mol NH3 = 4 mol NH3 1 mol N2 A second way: 6 mol H2 x 2 mol NH3 = 4 mol NH3 3 mol H2 4 moles of ammonia is all the ammonia we can get when we react 2 moles nitrogen and 6 moles hydrogen. This amount of product is called the theoretical yield.

Limiting Reactant N2 (g) + 3H2 (g)  2NH3 (g) If we start with 2 moles nitrogen and 7 moles hydrogen, how many moles of ammonia is it theoretically possible to produce? Still just 4 moles…we do not have enough N2 to make more than 4 moles of NH3. We call the reactant that is NOT in excess the limiting reactant. In this example, N2 is the limiting reactant. H2 is in excess. Limiting reactant problems are usually ones in which amounts of all reactants are given (as opposed to just one).

Limiting Reactant Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves both in an excess of water. What is the theoretical yield of Au(OH)3, if the equation for the reaction is the one given below? 2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O(l)  2Au(OH)3(s) + 6NaCl(aq) + 2KCl(aq) + 3CO2(g) This problem is worked with the same five steps as for the stoichiometry equations, BUT… you must perform the calculation for each reactant (unless it’s reported to be in excess)! Whichever reactant gives you LESS product is the limiting reactant.

Limiting Reactant = 13.13 g Au(OH)3 20.00 g 25.00 g ? g Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves both in an excess of water. What is the theoretical yield of Au(OH)3, if the equation for the reaction is the one given below? 20.00 g 25.00 g ? g 2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O(l)  2Au(OH)3(s) + 6NaCl(aq) + 2KCl(aq) + 3CO2(g) Calculate the theoretical yield of Au(OH)3 from 20.00 g KAuCl4: 20.00 g KAuCl4 x 1 mol KAuCl4 x 2 mol Au(OH)3 x 247.99 g Au(OH)3 377.88 g KAuCl4 2 mol KAuCl4 1 mol Au(OH)3 = 13.13 g Au(OH)3 The results of this calculation mean that if we had 20.00 g KAuCl4 and an excess of Na2CO3, we could produce 13.13 g of Au(OH)3.

Limiting Reactant = 39.00 g Au(OH)3 20.00 g 25.00 g ? g Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves both in an excess of water. What is the theoretical yield of Au(OH)3, if the equation for the reaction is the one given below? 20.00 g 25.00 g ? g 2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O(l)  2Au(OH)3(s) + 6NaCl(aq) + 2KCl(aq) + 3CO2(g) Calculate the theoretical yield of Au(OH)3 from 25.00 g Na2CO3: 25.00 g Na2CO3 x 1 mol Na2CO3 x 2 mol Au(OH)3 x 247.99 g Au(OH)3 105.99 g Na2CO3 3 mol Na2CO3 1 mol Au(OH)3 = 39.00 g Au(OH)3 The results of this calculation mean that if we had 25.00 g Na2CO3 and an excess of KAuCl4, we could produce 39.00 g of Au(OH)3.

Limiting Reactant Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves both in an excess of water. What is the theoretical yield of Au(OH)3, if the equation for the reaction is the one given below? 20.00 g 25.00 g ? g 2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O(l)  2Au(OH)3(s) + 6NaCl(aq) + 2KCl(aq) + 3CO2(g) If we had 20.00 g KAuCl4 and an excess of Na2CO3, we could produce 13.13 g of Au(OH)3. If we had 25.00 g Na2CO3 and an excess of KAuCl4, we could produce 39.00 g of Au(OH)3. 13.13 g is the smaller amount of product, which means KAuCl4 is the limiting reactant, and the theoretical yield is 13.13 g.

Limiting Reactant = 8.415 g Na2CO3 20.00 g ? g 13.13 g Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves both in an excess of water. How much of the excess reactant is present at the completion of the reaction? 20.00 g ? g 13.13 g 2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O(l)  2Au(OH)3(s) + 6NaCl(aq) + 2KCl(aq) + 3CO2(g) First method: Use the limiting reactant to calculate how much of the excess reactant is actually needed in the reaction. 20.00 g KAuCl4 x 1 mol KAuCl4 x 3 mol Na2CO3 x 105.99 g Na2CO3 377.88 g KAuCl4 2 mol KAuCl4 1 mol Na2CO3 = 8.415 g Na2CO3 The results of this calculation mean 8.415 g Na2CO3 is the mass needed to react with 20.00 g KAuCl4. Any mass beyond that is excess: Excess Na2CO3 = 25.00 g – 8.415 g = 16.58 g

Limiting Reactant = 8.418 g Na2CO3 20.00 g ? g 13.13 g Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves both in an excess of water. How much of the excess reactant is present at the completion of the reaction? 20.00 g ? g 13.13 g 2KAuCl4(aq) + 3Na2CO3(aq) + 3H2O(l)  2Au(OH)3(s) + 6NaCl(aq) + 2KCl(aq) + 3CO2(g) Second method: Use the theoretical yield to calculate how much of the excess reactant is actually needed in the reaction. 13.13 g Au(OH)3 x 1 mol Au(OH)3 x 3 mol Na2CO3 x 105.99 g Na2CO3 247.99 g Au(OH)3 2 mol Au(OH)3 1 mol Na2CO3 = 8.418 g Na2CO3 The results of this calculation mean 8.418 g Na2CO3 is the mass needed to produce 13.13 g Au(OH)3. Any mass beyond that is excess: Excess Na2CO3 = 25.00 g – 8.418 g = 16.58 g

Unit 4 Lecture 4 - Limiting Reactants Limiting Reactant – Percent Yield Often reactions do not go to 100% completion. Percent yield gives a quantitative measure of the extent to which the reaction went to completion. Percent yield = 100 x actual yield theoretical yield Example: A chemist takes 20.00 g KAuCl4 and 25.00 g Na2CO3 and dissolves both in an excess of water. The Au(OH)3 recovered is found to have a mass of 10.83 g. What is the percent yield of the reaction? We already have calculated the theoretical yield and found it to be 13.13g. % yield = 100 x 10.83 g = 82.48 % 13.13 g

Unit 4 Lecture 4 - Limiting Reactants One more limiting reactant problem The reaction below is run starting with 1.000 g nitrogen and 2.000 g lithium. Report the grams of each species present at the end of the reaction if a) the yield is 100% and b) the yield is 75%. 6Li(s) + N2 (g)  2Li3N(s) 100%: Li = 0.513 g, N2 = 0 g, Li3N = 2.487 g 75%: Li = 0.885 g, N2 = 0.250 g, Li3N = 1.865 g

Unit 4 Lecture 4 - Limiting Reactants More problems How many kilograms of aluminum are required to produce 2.00 kilograms of hydrogen according to the following reaction? 2Al(s) + 6H+(aq)  3H2(g) + 2Al3+(aq) 17.8 kg

Unit 4 Lecture 4 - Limiting Reactants More problems One molecule of penicillin G has a mass of 5.553 x 10-22 g. What is the molar mass of penicillin G? 334.4 g Hemoglobin has four iron atoms per molecule and contains 0.340% iron by mass. Calculate the molar mass of hemoglobin. 65700 g

Unit 4 Lecture 4 - Limiting Reactants More problems Serotonin has 68.2 wt% C, 6.86 wt% H, 15.9 wt% N, and 9.08 wt% O. Its molar mass is 176 g. Write its molecular formula. C10H12N2O

Unit 4 Lecture 4 - Limiting Reactants Name or write formula HNO2 permanganic acid H2SO4 iron(II) sulfate HF(aq) hypochlorous acid HC2H3O2 zinc nitrate FeP silver phosphate CuOH ammonium carbonate K2CO3 lithium bromide NaHCO3 sulfur dioxide NH3 sodium sulfite magnesium hydrogen sulfate