2. Section 3.7 Limiting Reactants

3. Grilled Cheese Sandwich Bread + Cheese  ‘Cheese Melt’ 2 B + C  B 2 C 100 bread 30 slices ? sandwiches What is the limiting factor in our ability to make the maximum about of grilled cheese sandwiches containing 2 slices of bread and 1 slice of cheese?

4. LIMITING REACTANT IMPORTANCE: Calculations of limiting reactant bring quantitative understanding to chemical reactions These calculations are used in both General and Organic Chemistry

5. DEFINITIONS LIMITING REACTANT Completely consumed in a chemical reaction Determines the amount of product formed The reactant that produces the least amount of product

6. DEFINITIONS THEORETICAL YIELD The amount of product that can be made based on the amount of the limiting reactant ACTUAL YIELD The amount of product actually or experimentally produced THE PERCENT YIELD %yield = (actual/theoretical) x 100

7. Limiting Reactants. An analogous situation occurs with chemical reactions. Consider the reaction: 2 H 2 (g)+O 2 (g)→2 H 2 O(l) 2 mol +1 mol2 mol If we have exactly 2 mol of H 2 and 1 mol of O 2, then we can make 2 mol of water. But what if we have 4 mol of H 2 and 1 mol of O 2. Now we can make only 2 mol H 2 O with 2 mol H 2 left over. In this case the O 2 is the limiting reagent. The limiting reagent is the one with nothing left over.

8. Container 1 Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 269

9. Before and After Reaction 1 All the hydrogen and nitrogen atoms combine. Before the reaction After the reaction

10. Container 2 Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 270

11. Before and After Reaction 2 Before the reactionAfter the reaction

12. Multiplying an equation through by a common multiple: We can multiply all the coefficients in a balanced equation by any multiple, and it still has the correct ratios of moles. Thus, if we have: Zn(s) + 2HCl(aq)  ZnCl 2(aq) +H 2 (g) 1 mole 2 moles 1 mole + 1 mole If we have 2 moles of Zn(s), this gives: (x 2) 2 moles 4 moles 2 moles 2 moles or if we have 0.5 moles Zn(s) we have: (x 0.5) 0.5 moles 1 mole 0.5 moles 0.5 moles

13. METHODS USED TO DETERMINE THE LIMITING REACTANT I.Calculate the moles needed of each reactant and compare with the moles given II. Divide the moles of each reactant by its stoichiometric coefficient and then compare them III.Calculate the moles of product produced by each reactant and compare them

14. Example I. Consider the reaction of H 2 and N 2 to give NH 3, and assume we have 3.0 mol N 2 and 6.0 mol H 2. We have the balanced equation: N 2 (g)+3 H 2 (g)→2 NH 3 (g) 1 mol3 mol2 mol Factor = moles N 2 we have moles N 2 in equation = 3.0 mol N 2 1.0 mol N 2 = 3.0 (multiply all coefficients in balanced equation by this factor)

15. Multiply all coefficients by factor (x 3): N 2 (g)+3 H 2 (g)→2 NH 3 (g) 1 mol 3 mol 2 mol 3 mol3 x 3 = 9 mol 3 x 2 = 6 mol Try N 2 as limiting reagent: 3 mol N 2 requires how many moles H 2 ? =3 x 3 = 9 mol We only have 6 mol H 2, so H 2 is the limiting reagent.

16. Example II. Divide the moles of each reactant by its stoichiometric coefficient Consider the following reaction: 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  Ba 3 (PO 4 ) 2 + 6 NaNO3 How much Ba 3 (PO 4 ) 2 can be formed if we have in the solutions 3.50 g sodium phosphate and 6.40 g barium nitrate?

17. Step 1. Convert to moles: First work out numbers of Moles: Na 3 PO 4 = 3.50 g x 1 mol = 0.0213 mol 164 g Ba(NO 3 ) 2 = 6.40 g x 1 mol= 0.0245 mol 261 g

18. Step 2. Divide moles by its stoichiometric coefficient 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  Ba 3 (PO 4 ) 2 + 6 NaNO3 Na 3 PO 4 : 0.0213 mol = 0.01065 2 mol Ba(NO 3 ) 2 : 0.0245 mol =0.00817 LR 3 mol

19. Example III. Calculate the amount of product produced by each reactant 1N 2 (g) + 3H 2 (g) → 2NH 3 (g) Given 3.0 mole 6.0 mole 3.0 mol N 2 x 2 mol NH 3 = 6.0 mol NH 3 1 mol N 2 6.0 mol H 2 x 2 mol NH 3 = 4.0 mol NH 3 (theoretical yield) 3 mol H 2 The reactant that produces the least amount of product is the L.R.  H 2

20. Practice Exercise: Zn metal (2.00 g) plus solution of AgNO 3 (2.50 g) reacts according to: Zn(s) +2 AgNO 3 (aq)  Zn(NO 3 ) 2 + 2 Ag(s) 1 mol2 mol Which is the limiting reagent? How much Zn will be left over?

21. Step 1. Convert to moles: Zn = 65 g/mol AgNO 3 = 108 + 14 + (3 x 16) = 170 g/mol Zn = 2.0 g x1 mol= 0.0308 mol 65 g AgNO 3 = 2.50 g x 1 mol =0.0147 mol 170 g

22. Step 2. Guess limiting reagent Zn(s) +2 AgNO 3 (aq)  Zn(NO 3 ) 2 + 2 Ag(s) 1 mol2 mol 0.0308 0.0147 In this case it seems clear that AgNO 3 must be the limiting reagent, because the equation says we must have 2 mols of AgNO 3 for each mol of Zn(s), but in fact we have more moles of Zn(s).

23. We can check this by dividing the moles of each reactant by their coefficients AgNO 3 = 0.0147/2 = 0.00735 Zn = 0.0308/1 = 0.0308 Zn(s) +2 AgNO 3 (aq)  Zn(NO 3 ) 2 + 2 Ag(s) 1 mol2 mol 0.0308 0.00735 We in fact have 0.0305 mol of Zn, which is more than the 0.00735 mol of AgNO 3, so AgNO 3 is clearly the limiting reactant.

24. How much Zn is left over? Use the limiting reactant to determine this: 0.0147 mol AgNO 3 x 1 mol Zn x 65 g Zn 2 mol AgNO 3 1 mol Zn = 0.47775 g Zn Subtract this from the amount of Zn available: 2.00 g Zn - 0.4775g Zn = 1.52 g Zn in excess

25. Homework # 3.71-3.74 on pages 115-116

26. Percent Yield: Theoretical yields: The quantity of product that forms if all of the limiting reagent reacts is called the theoretical yield. Usually, we obtain less than this, which is known as the actual yield. Percent yield = actual yield x 100 Theoretical yield

27. Problem: 10.4 g of Ba(OH) 2 was reacted with an excess of Na 2 SO 4 to give a precipitate of BaSO 4. If the reaction actually yielded 11.2 g of BaSO 4, what is a) the theoretical yield of BaSO 4 and b) what is the percentage yield of BaSO 4 ? The balanced equation for the reaction is: Ba(OH) 2 (aq) + Na 2 SO 4 (aq)  BaSO 4 (s) + 2 NaOH(aq)

28. Step 1. Convert to moles: Ba(OH) 2 (aq) + Na 2 SO 4 (aq)  BaSO 4 (s) + 2 NaOH(aq) 1 mole1 mole 1 mole 2 moles Moles Ba(OH) 2 : Mol. Mass Ba(OH) 2 = 137.3 + 2 x (16.0 + 1.0) = 171.3 g/mol Moles = 10.4 g x 1 mol =0.0607 moles 171.3 g

29. Step 2. Work out how much BaSO 4 will be formed: Ba(OH) 2 (aq) + Na 2 SO 4 (aq)  BaSO 4 (s) + 2 NaOH(aq) 1 mole1 mole 1 mole 2 moles 0.0607 moles When it says that one reagent is in excess, that means we do not have to worry about that reagent, and the other one is the limiting reagent, in this case the BaSO 4. We see that 1 mole of Ba(OH) 2 will produce 1 mole of BaSO 4. Our factor is thus 0.0607, and we will get 0.0607 moles of BaSO 4.

30. Convert actual yield to percentage yield: Percent yield =actual yield x 100 % Theoretical yield =11.2 g x100 % 14.29 g =78.4 % yield