1. Lecture 6. 3.6: Quantitative Information from Balanced Equations. 2 H 2 + (1) O 2 → 2 H 2 O 2 moles 1 mole 2 moles

2. 3.6: Quantitative Information from Balanced Equations. The coefficients in a balanced equation represent both the relative numbers of molecules involved in the reaction, AND THE RELATIVE NUMBERS OF MOLES, and therefore the relative masses: 2 H 2 (g)+O 2 (g) →2 H 2 O (l) 2 molecules + 1molecule2 molecules 2 moles 1 mole 2 moles = 2 x (2.0) g + 32.0 g→2 x (18.0) g

3. Coefficients in a balanced equation: The coefficients in a balanced equation as written mean ‘molecules’. So we have 2 H 2 (g)+O 2 (g) →2 H 2 O (l) 2 molecules + 1molecule2 molecules We can multiply through the whole equation by Avogadro’s number, and then we have moles: 2 x 6.022 x 10 23 1 x 6.022 x 10 23 2 x 6.022 x 10 23 = 2 moles 1 mole 2 moles

4. Weights of products and reactants in a balanced equation: We can thus work out how many grams of water will be produced by burning given amounts of H 2 and O 2 together. Example: The combustion of butane: 2 C 4 H 10 (l) + 13 O 2 (g) → 8 CO 2 (g) + 10 H 2 O(l) 2 moles 13 moles 8 moles 10 moles 2 x 58.0 g 8 x 44.0 g M. Wt. C 4 H 10 M. Wt. CO 2

5. How many grams of CO 2 will be obtained by burning 1.00 g of C 4 H 10 ? 1) Grams reactant → moles Conversion factor: 1 = 1 mol/58.0 g 1.00 g x 1mol/58 g = 0.0172 mol 2) Moles reactant → Moles product 2 moles C 4 H 10 → 8 moles CO 2 moles CO 2 = 0.0172 x 8/2 = 0.0688 moles 3) Moles product → grams product 0.0688 moles CO 2 = 0.0688 mol x 44.0 g/1 mol grams of CO 2 = 3.03 g coefficients from balanced equation

6. 3.7. Limiting Reactants. Cheese sandwiches: (Ch = slice of cheese, Bd = slice of bread) 2 Bd + 1 Ch=Bd 2 Ch If we have 12 Ch and 8 Bd, how many sandwiches can we make? Obviously, only 4, with 8 Ch left over. We are limited in this case by the amount of Bd we have. Bd in this case is the limiting ingredient.

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. 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

9. How do we solve limiting reagent problems? We try both reagents out as the possible limiting reagent. 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. Which is the limiting reagent? 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)

10. 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.

11. Example: Consider the following reaction: 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  2 mol 3 mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO3(aq) 1 mol 6 mol 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?

12. 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

13. Step 2. Guess limiting reagent (it doesn’t matter if you guess wrong): Guess that Ba(NO 3 ) 2 is the limiting reagent: 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  2 mol3 mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) 1 mol6 mol Factor =0.0245 mol (moles we have) 3 mol(moles in equation) =0.00817

14. Guessing which reagent is the limiting reagent: 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  2 mol3 mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) 1 mol6 mol We notice that we have only slightly more moles of Ba(NO 3 ) 2 (0.0245 moles) than Na 3 PO 4 (0.0213 moles), but the Ba(NO 3 ) 2 has a coefficient of 3 in the balanced equation, while that of Na 3 PO 4 is only 2. In general go for the substance with the higher coefficient and higher molecular mass.

15. Step 3. Multiply all the moles in the equation by the factor: 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  2 mol3 mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) 1 mol6 mol times factor (0.00817): 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  2 mol3 mol or 0.01634 mol 0.0245 mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) 1 mol6 mol 0.00817 mol0.0490 mol

16. Step 4. Compare required moles of Na 3 PO 4 with moles we previously calculated (0.0213 moles) 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq) = 0.01634 mol 0.0245 mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) 0.00817 mol0.0490 mol We have 0.0213 moles of Na 3 PO 4, so we guessed right, Ba(NO 3 ) 2 is limiting reagent (if we guessed wrong calculate new factor with other reagent and repeat calculations). bigger than

17. How much Ba 3 (PO 4 ) 2 can be formed? We now have an equation with the correct number of moles of each reactant if the limiting reactant is 0.0245 moles of Ba(NO 3 ) 2 : 2 Na 3 PO 4 (aq) + 3 Ba(NO 3 ) 2 (aq)  0.01634 mol 0.0245 mol Ba 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq) 0.00817 mol0.0490 mol So we need to calculate how many grams of Ba 3 (PO 4 ) 2 there are in 0.00817 moles.

18. Step 5. Convert moles of Ba 3 (PO 4 ) 2 to grams: We now need g of Ba 3 (PO 4 ) 2. F. Wt. = 3 x 137.3 + 2 x 31.0 + 8 x 16.0 = 601.9 g/mol Need grams, so we have 0.00817 mol x601.9 g= 4.92 g. 1 mol

19. 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?

20. Step 1. Convert to moles: Zn = 65.39 g/mol AgNO 3 = 107.9 + 14 + (3 x 16) = 169.9 g/mol Zn = 2.0 g x1 mol= 0.0305 mol 65.39 g AgNO 3 = 2.50 g x 1 mol =0.0147 mol 169.9 g

21. Step 2. Guess limiting reagent Zn(s) +2 AgNO 3 (aq)  Zn(NO 3 ) 2 + 2 Ag(s) 1 mol2 mol 0.0305 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).

22. We can check this by using the AgNO 3 to calculate a factor that can be used to multiply through the equation. Factor = 0.0147/2 = 0.00735 Zn(s) +2 AgNO 3 (aq)  Zn(NO 3 ) 2 + 2 Ag(s) 1 mol2 mol 0.00735 0.0147 We in fact have 0.0305 mol of Zn, which is more than the 0.00735 mol required, so AgNO 3 is clearly the limiting reactant.

23. How much Zn will be left over? We actually have 0.0305 mol of Zn, but require only 0.00735 mol. We will therefore have 0.0305 mol – 0.00735 mol = 0.02315 mol = 65.4 x 0.02315 mol = 1.52 g left over (At. Wt. Zn)

24. 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 yieldx 100 Theoretical yield

25. 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)

26. 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

27. 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.

28. Calculation of theoretical yield: Convert 0.0607 moles of BaSO 4 to grams: Formula mass BaSO 4 = 137.3 + 32.0 = 4 x 16.0 =235.3 g/mol No of grams of BaSO 4 expected is: 0.0607 moles x 235.3 g =14.29 g 1 mole (theoretical yield)

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