Section 3.7

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Section 3.7

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1. Section 3.7 Limiting Reactants

2. Grilled Cheese Sandwich

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

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

5. 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

6. Limiting Reactants. An analogous situation occurs with chemical reactions. Consider the reaction: 2 H2(g) + O2(g) ? 2 H2O(l) 2 mol + 1 mol 2 mol If we have exactly 2 mol of H2 and 1 mol of O2, then we can make 2 mol of water. But what if we have 4 mol of H2 and 1 mol of O2. Now we can make only 2 mol H2O with 2 mol H2 left over. In this case the O2 is the limiting reagent. The limiting reagent is the one with nothing left over.

7. Container 1

8. Before and After Reaction 1

9. Container 2

10. Before and After Reaction 2

11. 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) ? ZnCl2(aq) + H2(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

12. METHODS USED TO DETERMINE THE LIMITING REACTANT 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 Calculate the moles of product produced by each reactant and compare them

13. Example I. Consider the reaction of H2 and N2 to give NH3, and assume we have 3.0 mol N2 and 6.0 mol H2. We have the balanced equation: N2(g) + 3 H2(g) ? 2 NH3(g) 1 mol 3 mol 2 mol Factor = moles N2 we have moles N2 in equation = 3.0 mol N2 1.0 mol N2 = 3.0 (multiply all coefficients in balanced equation by this factor)

14. Multiply all coefficients by factor (x 3): N2(g) + 3 H2(g) ? 2 NH3(g) 1 mol 3 mol 2 mol 3 mol 3 x 3 = 9 mol 3 x 2 = 6 mol Try N2 as limiting reagent: 3 mol N2 requires how many moles H2? = 3 x 3 = 9 mol We only have 6 mol H2, so H2 is the limiting reagent.

15. Example II. Divide the moles of each reactant by its stoichiometric coefficient Consider the following reaction: 2 Na3PO4(aq) + 3 Ba(NO3)2(aq) ?Ba3(PO4)2 + 6 NaNO3 How much Ba3(PO4)2 can be formed if we have in the solutions 3.50 g sodium phosphate and 6.40 g barium nitrate?

16. Step 1. Convert to moles: First work out numbers of Moles: Na3PO4 = 3.50 g x 1 mol = 0.0213 mol 164 g Ba(NO3)2 = 6.40 g x 1 mol = 0.0245 mol 261 g

17. Step 2. Divide moles by its stoichiometric coefficient 2 Na3PO4(aq) + 3 Ba(NO3)2(aq) ?Ba3(PO4)2 + 6 NaNO3 Na3PO4 : 0.0213 mol = 0.01065 2 mol Ba(NO3)2 : 0.0245 mol = 0.00817 LR 3 mol

18. Example III. Calculate the amount of product produced by each reactant 1N2(g) + 3H2(g) ? 2NH3(g) Given 3.0 mole 6.0 mole 3.0 mol N2 x 2 mol NH3 = 6.0 mol NH3 1 mol N2 6.0 mol H2 x 2 mol NH3 = 4.0 mol NH3 (theoretical yield) 3 mol H2 The reactant that produces the least amount of product is the L.R.?H2

19. Practice Exercise: Zn metal (2.00 g) plus solution of AgNO3 (2.50 g) reacts according to: Zn(s) + 2 AgNO3(aq) ? Zn(NO3)2+ 2 Ag(s) 1 mol 2 mol Which is the limiting reagent? How much Zn will be left over?

20. Step 1. Convert to moles: Zn = 65 g/mol AgNO3 = 108 + 14 + (3 x 16) = 170 g/mol Zn = 2.0 g x 1 mol = 0.0308 mol 65 g AgNO3 = 2.50 g x 1 mol = 0.0147 mol 170 g

21. Step 2. Guess limiting reagent Zn(s) + 2 AgNO3(aq) ? Zn(NO3)2+ 2 Ag(s) 1 mol 2 mol 0.0308 0.0147 In this case it seems clear that AgNO3 must be the limiting reagent, because the equation says we must have 2 mols of AgNO3 for each mol of Zn(s), but in fact we have more moles of Zn(s).

22. We can check this by dividing the moles of each reactant by their coefficients AgNO3 = 0.0147/2 = 0.00735 Zn = 0.0308/1 = 0.0308 Zn(s) + 2 AgNO3(aq) ? Zn(NO3)2+ 2 Ag(s) 1 mol 2 mol 0.0308 0.00735 We in fact have 0.0305 mol of Zn, which is more than the 0.00735 mol of AgNO3, so AgNO3 is clearly the limiting reactant.

23. How much Zn is left over? Use the limiting reactant to determine this: 0.0147 mol AgNO3 x 1 mol Zn x 65 g Zn 2 mol AgNO3 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

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

25. 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

26. Problem: 10.4 g of Ba(OH)2 was reacted with an excess of Na2SO4 to give a precipitate of BaSO4. If the reaction actually yielded 11.2 g of BaSO4, what is a) the theoretical yield of BaSO4 and b) what is the percentage yield of BaSO4? The balanced equation for the reaction is: Ba(OH)2(aq) + Na2SO4(aq) ? BaSO4(s) + 2 NaOH(aq)

27. Step 1. Convert to moles: Ba(OH)2(aq) + Na2SO4(aq) ? BaSO4(s) + 2 NaOH(aq) 1 mole 1 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

28. Step 2. Work out how much BaSO4 will be formed: Ba(OH)2(aq) + Na2SO4(aq) ? BaSO4(s) + 2 NaOH(aq) 1 mole 1 mole 1 mole 2 moles 0.0607 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 BaSO4. We see that 1 mole of Ba(OH)2 will produce 1 mole of BaSO4. Our factor is thus 0.0607, and we will get 0.0607 moles of BaSO4.

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

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