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UNIT 4 Quantitative Information from Balanced Chemical Equations.

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Presentation on theme: "UNIT 4 Quantitative Information from Balanced Chemical Equations."— Presentation transcript:

1 UNIT 4 Quantitative Information from Balanced Chemical Equations

2 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 C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (l) 2. Write what you know. Write what you need to know.

3 3.62 mol ? mol C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (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 mol C 3 H 8 x 5 mol O 2 = 18.1 mol O 2 1 mol C 3 H 8 stoichiometric equivalence (from chemical equation) Here we were asked for moles, so no mole/mass conversion was necessary. Quantitative Information from Balanced Equations

4 Example: What mass of oxygen is consumed in the combustion of g of propane? 1. Write and balance the equation for the reaction g ? g C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (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 g C 3 H 8 x 1 mol C 3 H 8 x 5 mol O 2 x g O 2 = g O g 1 mol C 3 H 8 1 mol O 2 Quantitative Information from Balanced Equations mass  moles stoichiometric equivalencemoles  mass

5 Detonation of nitroglycerin proceeds as follows: 4C 3 H 5 N 3 O 9 (l)  12CO 2 (g) + 6N 2 (g) + O 2 (g) + 10H 2 O(g) a) If a sample containing 2.00 mL of nitroglycerin (density = 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 N 2 are produced in the detonation? answers: a) mol, b) 5.6 L, c) g Quantitative Information from Balanced Equations - Example

6 N 2 (g) + 3H 2 (g)  2NH 3 (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 N 2 x 2 mol NH 3 = 4 mol NH 3 1 mol N 2 A second way: 6 mol H 2 x 2 mol NH 3 = 4 mol NH 3 3 mol H 2 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

7 N 2 (g) + 3H 2 (g)  2NH 3 (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 N 2 to make more than 4 moles of NH 3. We call the reactant that is NOT in excess the limiting reactant. In this example, N 2 is the limiting reactant. H 2 is in excess. Limiting reactant problems are usually ones in which amounts of all reactants are given (as opposed to just one). Limiting Reactant

8 Example: A chemist takes g KAuCl 4 and g Na 2 CO 3 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? 2KAuCl 4 (aq) + 3Na 2 CO 3 (aq) + 3H 2 O(l)  2Au(OH) 3 (s) + 6NaCl(aq) + 2KCl(aq) + 3CO 2 (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

9 Example: A chemist takes g KAuCl 4 and g Na 2 CO 3 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? g25.00 g? g 2KAuCl 4 (aq) + 3Na 2 CO 3 (aq) + 3H 2 O(l)  2Au(OH) 3 (s) + 6NaCl(aq) + 2KCl(aq) + 3CO 2 (g) Calculate the theoretical yield of Au(OH) 3 from g KAuCl 4 : g KAuCl 4 x 1 mol KAuCl 4 x 2 mol Au(OH) 3 x g Au(OH) g KAuCl 4 2 mol KAuCl 4 1 mol Au(OH) 3 = g Au(OH) 3 The results of this calculation mean that if we had g KAuCl 4 and an excess of Na 2 CO 3, we could produce g of Au(OH) 3. Limiting Reactant

10 Example: A chemist takes g KAuCl 4 and g Na 2 CO 3 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? g25.00 g? g 2KAuCl 4 (aq) + 3Na 2 CO 3 (aq) + 3H 2 O(l)  2Au(OH) 3 (s) + 6NaCl(aq) + 2KCl(aq) + 3CO 2 (g) Calculate the theoretical yield of Au(OH) 3 from g Na 2 CO 3 : g Na 2 CO 3 x 1 mol Na 2 CO 3 x 2 mol Au(OH) 3 x g Au(OH) g Na 2 CO 3 3 mol Na 2 CO 3 1 mol Au(OH) 3 = g Au(OH) 3 The results of this calculation mean that if we had g Na 2 CO 3 and an excess of KAuCl 4, we could produce g of Au(OH) 3. Limiting Reactant

11 Example: A chemist takes g KAuCl 4 and g Na 2 CO 3 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? g25.00 g? g 2KAuCl 4 (aq) + 3Na 2 CO 3 (aq) + 3H 2 O(l)  2Au(OH) 3 (s) + 6NaCl(aq) + 2KCl(aq) + 3CO 2 (g) If we had g KAuCl 4 and an excess of Na 2 CO 3, we could produce g of Au(OH) 3. If we had g Na 2 CO 3 and an excess of KAuCl 4, we could produce g of Au(OH) g is the smaller amount of product, which means KAuCl 4 is the limiting reactant, and the theoretical yield is g. Limiting Reactant

12 Example: A chemist takes g KAuCl 4 and g Na 2 CO 3 and dissolves both in an excess of water. How much of the excess reactant is present at the completion of the reaction? g ? g13.13 g 2KAuCl 4 (aq) + 3Na 2 CO 3 (aq) + 3H 2 O(l)  2Au(OH) 3 (s) + 6NaCl(aq) + 2KCl(aq) + 3CO 2 (g) First method: Use the limiting reactant to calculate how much of the excess reactant is actually needed in the reaction g KAuCl 4 x 1 mol KAuCl 4 x 3 mol Na 2 CO 3 x g Na 2 CO g KAuCl 4 2 mol KAuCl 4 1 mol Na 2 CO 3 = g Na 2 CO 3 The results of this calculation mean g Na 2 CO 3 is the mass needed to react with g KAuCl 4. Any mass beyond that is excess: Excess Na 2 CO 3 = g – g = g Limiting Reactant

13 Example: A chemist takes g KAuCl 4 and g Na 2 CO 3 and dissolves both in an excess of water. How much of the excess reactant is present at the completion of the reaction? g ? g13.13 g 2KAuCl 4 (aq) + 3Na 2 CO 3 (aq) + 3H 2 O(l)  2Au(OH) 3 (s) + 6NaCl(aq) + 2KCl(aq) + 3CO 2 (g) Second method: Use the theoretical yield to calculate how much of the excess reactant is actually needed in the reaction g Au(OH) 3 x 1 mol Au(OH) 3 x 3 mol Na 2 CO 3 x g Na 2 CO g Au(OH) 3 2 mol Au(OH) 3 1 mol Na 2 CO 3 = g Na 2 CO 3 The results of this calculation mean g Na 2 CO 3 is the mass needed to produce g Au(OH) 3. Any mass beyond that is excess: Excess Na 2 CO 3 = g – g = g Limiting Reactant

14 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 g KAuCl 4 and g Na 2 CO 3 and dissolves both in an excess of water. The Au(OH) 3 recovered is found to have a mass of 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 g = % g Limiting Reactant – Percent Yield

15 The reaction below is run starting with g nitrogen and 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) + N 2 (g)  2Li 3 N(s) 100%: Li = g, N 2 = 0 g, Li 3 N = g 75%: Li = g, N 2 = g, Li 3 N = g One more limiting reactant problem

16 How many kilograms of aluminum are required to produce 2.00 kilograms of hydrogen according to the following reaction? 2Al(s) + 6H + (aq)  3H 2 (g) + 2Al 3+ (aq) 17.8 kg More problems

17 One molecule of penicillin G has a mass of x g. What is the molar mass of penicillin G? g Hemoglobin has four iron atoms per molecule and contains 0.340% iron by mass. Calculate the molar mass of hemoglobin g More problems

18 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. C 10 H 12 N 2 O More problems

19 HNO 2 permanganic acid H 2 SO 4 iron(II) sulfate HF(aq)hypochlorous acid HC 2 H 3 O 2 zinc nitrate FePsilver phosphate CuOHammonium carbonate K 2 CO 3 lithium bromide NaHCO 3 sulfur dioxide NH 3 sodium sulfite magnesium hydrogen sulfate Name or write formula


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