Limiting Reactants. Copyright 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Transcription

1 Limiting Reactants Copyright 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

2 Limiting Reactants Limiting Reactant used up in a reaction Determines/limits the amount of product Stops the reaction Ecess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle Courtesy Christy Johannesson

3 Reacting Amounts In a table setting, there is 1 plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from 5 plates, 6 forks, 4 spoons, and 7 knives? Copyright 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings What is the limiting item?

4 Reacting Amounts Four table settings can be made. Initially Use Left over plates forks spoons knives The limiting item is the spoon.

5 Eample of Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 jar of peanut butter? With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item. Copyright 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

6 Eample of Everyday Limiting Reactant How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter? With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item. Copyright 2008 by Pearson Education, Inc. publishing as Benjamin Cummings

7 Limiting Reactants When 4.00 mol H 2 is mied with 2.00 mol Cl 2,how many moles of HCl can form? H 2 (g) + Cl 2 (g) 2HCl (g) 4.00 mol 2.00 mol??? Mol Calculate the moles of product from each reactant, H 2 and Cl 2. The limiting reactant is the one that produces the smaller amount of product.

8 Limiting Reactants aluminum + chlorine gas aluminum chloride Al(s) + Cl 2 (g) AlCl 3 2 Al(s) + 3 Cl 2 (g) 2 AlCl g 100 g? g A. 200 g B. 125 g C. 667 g D. 494 g

9 Limiting Reactants aluminum + chlorine gas aluminum chloride 2 Al(s) + 3 Cl 2 (g) 2 AlCl g 100. g g How much product would be made if we begin with 100 g of aluminum? g AlCl 3 = 100 g Al 1 mol Al 2 mol AlCl 3 27 g Al 2 mol Al g AlCl 3 1 mol AlCl 3 = 494 g AlCl 3 Al AlCl 3 How much product would be made if we begin with 100 g of chlorine gas? 1 mol Cl 2 2 mol AlCl g AlCl 3 g AlCl 3 = 100 g Cl 2 71 g Cl 2 = 125 g AlCl 3 3 mol Cl 2 1 mol AlCl 3 Cl 2 AlCl 3

10 Limiting Reactants Using Mass If 4.80 mol Ca mied with 2.00 mol N 2, which is the limiting reactant? 3Ca(s) + N 2 (g) Ca 3 N 2 (s) Moles of Ca 3 H 2 from Ca 4.80 mol Ca 1 mol Ca 3 N 2 = 1.60 mol Ca 3 N 2 3 mol Ca (Ca used up) Moles of Ca 3 H 2 from N mol N 2 1 mol Ca 3 N 2 = 2.00 mol Ca 3 N 2 1 mol N 2 (not possible) All Ca is used up when 1.60 mol Ca 3 N 2 forms. Thus, Ca is the limiting reactant.

11 Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H 2 and 24.0 g O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l)

12 Limiting Reactants Using Mass Calculate the grams of H 2 for each reactant. H 2 : 8.00 g H 2 1 mol H 2 2 mol H 2 O g H 2 O g H 2 2 mol H 2 1 mol H 2 O = 71.5 g H 2 O (not possible) O 2 : 24.0 g O 2 1 mol O 2 2 mol H 2 O g H 2 O g O 2 1 mol O 2 1 mol H 2 O = 27.0 g H 2 O (smaller) O 2 is the limiting reactant.

13 Eample: Find the limiting reagent when 1.22g O 2 reacts with 1.05g H 2 to produce H 2 O. There are two solution methods you could use. In both methods, the first step is to convert the mass to moles.

14 Method 1 Use the moles of one reactant to calculate the necessary moles of the other reactant to fully react. Compare the calculated value with the actual value to see if this reagent is ecess or limiting.

15 Eample: Find the limiting reagent when 1.22g O 2 reacts with 1.05g H 2 to produce H 2 O. Answers using method 1: Convert mass to moles: mol O 2, 0.5 mol H 2 Calculate H 2 moles necessary to react with O 2 : mol H 2. Compare mol H 2 to actual mol of H 2 (0.5mol H 2 ), Since 0.5 mol H 2 is more than mol H 2, H 2 is the ecess reagent and O 2 is the limiting reagent.

16 Method 2 Use the moles of each of the reactant to calculate one of the products. The reagent that gave the smaller calculated value of product is the limiting reagent. The actual value of the amount of product is the smaller of the calculated values.

17 Eample: Find the limiting reagent when 1.22g O 2 reacts with 1.05g H 2 to produce H 2 O. Answers using Method 2: Convert mass to moles: mol O 2, 0.5 mol H 2 Calculate H 2 O moles produced by using each of the reactants: Using O 2 : mol H 2 O. Using H 2 : 0.5 mol H 2 O. The actual amount H 2 O produced is the smaller one of the two values(0.076mol H 2 O). O 2 is the limiting reagent, since O 2 was used in the calculation of the 0.076mol H 2 O.

18 Practice questions 1. 2Al + 6HCl 2AlCl 3 + 3H 2 If 25 g of aluminum was added to 90 g of HCl, what mass of H 2 will be produced (try this two ways with a chart & using the shortcut)? 2. N 2 + 3H 2 2NH 3 : If you have 20 g of N 2 and 5.0 g of H 2, which is the limiting reagent? 3. What mass of aluminum oide is formed when 10.0 g of Al is burned in 20.0 g of O 2? 4. When C 3 H 8 burns in oygen, CO 2 and H 2 O are produced. If 15.0 g of C 3 H 8 reacts with 60.0 g of O 2, how much CO 2 is produced? 5. How can you tell if a question is a limiting reagent question vs. typical stoichiometry?

19 1 #mol Al = 25 g Al 1 mol Al = mol 27.0g Al # mol HCl = 90 g HCl 1 mol HCl = mol 36.5g HCl What we have What we need # g H 2 = 90 g HCl Al HCl / /0.926 = 1 mol = 2.7 mol 2 6 2/2 = 1 mol 6/2 = 3 mol HCl is limiting. 1 mol HCl 3 mol H g H 2 = 2.47 g H g HCl 6 mol HCl 1 mol H 2

20 Question 1: shortcut 2Al + 6HCl 2AlCl 3 + 3H 2 If 25 g aluminum was added to 90 g HCl, what mass of H 2 will be produced? # g H 3 mol H 2 2 = 25 g Al 1 mol Al 2.0 g H 2 = 2.78 g H g Al 2 mol Al 1 mol H 2 # g H 3 mol H 2 2 =90 g HCl 1 mol HCl 2.0 g H 2 = 2.47 g H g HCl 6 mol HCl 1 mol H 2

21 Question 2: shortcut N 2 + 3H 2 2NH 3 If you have 20 g of N 2 and 5.0 g of H 2, which is the limiting reagent? # g NH 3 = 20 g N 1 mol N 2 mol NH g NH 2 3 = 24.3 g H g N 2 1 mol N 2 1 mol NH 3 # g NH 3 = 5.0 g H 1 mol H 2 mol NH g NH 2 = g H g H 2 3 mol H 2 1 mol NH 3 N 2 is the limiting reagent

22 Question 3: shortcut 4Al + 3O 2 2 Al 2 O 3 What mass of aluminum oide is formed when 10.0 g of Al is burned in 20.0 g of O 2? # g Al 2 O 3 = 10.0 g Al 1 mol Al 2 mol Al 2 O g Al O 2 3 = 18.9 g Al 2 O g Al 4 mol Al 1 mol H 2 # g Al 2 O 3 = 20.0 g O 1 mol O 2 mol Al 2 O g Al 2 O 3 = 42.5 g Al 2 O g O 2 3 mol O 2 1 mol H 2

23 Question 4: shortcut C 3 H 8 + 5O 2 3CO 2 + 4H 2 O When C 3 H 8 burns in oygen, CO 2 and H 2 O are produced. If 15.0 g of C 3 H 8 reacts with 60.0 g of O 2, how much CO 2 is produced? # g CO 2 = 15.0 g C 3 mol CO 2 3 H 1 mol C 44.0 g CO 3 H = 45.0 g CO g C 3 H 8 1 mol C 3 H 8 1 mol CO 2 # g CO 2 = 60.0 g O 1 mol O 3 mol CO g CO 2 = 49.5 g CO g O 2 5 mol O 2 1 mol CO 2 5. Limiting reagent questions give values for two or more reagents (not just one)

24 Question 2 # mol N 1 mol N 2 2 = 20 g N 2 = mol N 2 28 g N 2 # mol H 1 mol H 2 2 = 5.0 g H 2 = 2.5 mol H 2 2 g H 2 What we have What we need N 2 H mol 2.5 mol 0.714/0.714 = 1 mol 2.5/0.714 = 3.5 mol 1 mol 3 mol We have more H 2 than what we need, thus H 2 is in ecess and N 2 is the limiting factor.

25 3 4Al + 3O 2 2 Al 2 O 3 # mol Al = 10 g Al 1 mol Al 0.37 mol Al = 27 g Al # mol O 1 mol O 2 2 = 20 g O mol O 2 = 2 32 g O 2 There is Al O mol mol more What we than have 0.37/ /0.37 = 1 mol = 1.68 mol enough What we 4 mol 3 mol O 2 ; Al is need 4/4 = 1 mol 3/4 = 0.75 mol limiting # g Al 2 mol Al 2 O 3 2 O 3 = 0.37 mol Al 102 g Al 2 O 3 4 mol Al 1 mol Al 2 O 3 = g Al O

26 4 C 3 H 8 + 5O 2 3CO 2 + 4H 2 O # mol C 1 mol C 3 H 8 3 H 8 =15 g C 3 H 0.34 mol 8 = 44 g C 3 H 8 C 3 H 8 # mol O 1 mol O 2 2 = 60 g O mol O 2 = 2 32 g O 2 C 3 H 8 O 2 We have What we 0.34 mol mol more than 0.34/ /0.34 enough O have 2, = 1 mol = 5.5 mol C 3 H 8 is limiting Need 1 mol 5 mol # g CO 2 = 0.34 mol C3 H 8 3 mol CO 2 1 mol C 3 H 8 = 44 g CO 2 1 mol CO 2 45 g CO 2

27 Limiting Reagents: shortcut MgCl 2 + 2AgNO 3 Mg(NO 3 ) 2 + 2AgCl If 25 g magnesium chloride was added to 68 g silver nitrate, what mass of AgCl will be produced? # g AgCl= 25 g MgCl 2 1 mol MgCl g MgCl 2 2 mol AgCl 1 mol MgCl 2 = g AgCl 1 mol AgCl g AgCl # g AgCl= 68 g AgNO 3 For more lessons, visit 1 mol AgNO g AgNO 3 2 mol AgCl g AgCl 2 mol AgNO 3 1 mol AgCl = g AgCl