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Lecture 6. 3.6: Quantitative Information from Balanced Equations.

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. 3.6: Quantitative Information from Balanced Equations.

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Lecture 6. 3.6: Quantitative Information from Balanced Equations.

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  1. Lecture 6. 3.6: Quantitative Information from Balanced Equations. 2 H2 + (1) O2 → 2 H2O 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 H2(g) + O2(g) → 2 H2O (l) 2 molecules + 1molecule 2 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 H2(g) + O2(g) → 2 H2O (l) 2 molecules + 1molecule 2 molecules We can multiply through the whole equation by Avogadro’s number, and then we have moles: 2 x 6.022 x 1023 1 x 6.022 x 10232 x 6.022 x 1023 = 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 H2 and O2 together. Example: The combustion of butane: 2 C4H10(l) + 13 O2(g) → 8 CO2(g) + 10 H2O(l) 2 moles 13 moles 8 moles 10 moles 2 x 58.0 g 8 x 44.0 g M. Wt. C4H10 M. Wt. CO2

  5. How many grams of CO2 will be obtained by burning 1.00 g of C4H10? 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 C4H10 → 8 moles CO2 moles CO2 = 0.0172 x 8/2 = 0.0688 moles 3) Moles product → grams product 0.0688 moles CO2 = 0.0688 mol x 44.0 g/1 mol grams of CO2 = 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 = Bd2Ch 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 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.

  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)  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

  9. How do we solve limiting reagent problems? We try both reagents out as the possible limiting reagent. Consider the reaction of H2 and N2 to give NH3, and assume we have 3.0 mol N2 and 6.0 mol H2. Which is the limiting reagent? 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)

  10. 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 mol3 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.

  11. Example: Consider the following reaction: 2 Na3PO4(aq) + 3 Ba(NO3)2(aq)  2 mol 3 mol Ba3(PO4)2(s) + 6 NaNO3(aq) 1 mol 6 mol 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?

  12. 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

  13. Step 2. Guess limiting reagent (it doesn’t matter if you guess wrong): Guess that Ba(NO3)2 is the limiting reagent: 2 Na3PO4(aq) + 3 Ba(NO3)2(aq)  2 mol 3 mol Ba3(PO4)2(s) + 6 NaNO3(aq) 1 mol 6 mol Factor = 0.0245 mol (moles we have) 3 mol (moles in equation) = 0.00817

  14. Guessing which reagent is the limiting reagent: 2 Na3PO4(aq) + 3 Ba(NO3)2(aq)  2 mol 3 mol Ba3(PO4)2(s) + 6 NaNO3(aq) 1 mol 6 mol • We notice that we have only slightly more moles of Ba(NO3)2 (0.0245 moles) than Na3PO4 (0.0213 moles), but the Ba(NO3)2 has a coefficient of 3 in the balanced equation, while that of Na3PO4 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 Na3PO4(aq) + 3 Ba(NO3)2(aq)  2 mol 3 mol Ba3(PO4)2(s) + 6 NaNO3(aq) 1 mol 6 mol times factor (0.00817): 2 Na3PO4(aq) + 3 Ba(NO3)2(aq)  2 mol 3 mol or 0.01634 mol 0.0245 mol Ba3(PO4)2(s) + 6 NaNO3(aq) 1 mol 6 mol 0.00817 mol 0.0490 mol

  16. Step 4. Compare required moles of Na3PO4 with moles we previously calculated (0.0213 moles) 2 Na3PO4(aq) + 3 Ba(NO3)2(aq) = 0.01634 mol 0.0245 mol Ba3(PO4)2(s) + 6 NaNO3(aq) 0.00817 mol 0.0490 mol We have 0.0213 moles of Na3PO4, so we guessed right, Ba(NO3)2 is limiting reagent (if we guessed wrong calculate new factor with other reagent and repeat calculations). bigger than

  17. How much Ba3(PO4)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(NO3)2: 2 Na3PO4(aq) + 3 Ba(NO3)2(aq)  0.01634 mol 0.0245 mol Ba3(PO4)2(s) + 6 NaNO3(aq) 0.00817 mol 0.0490 mol So we need to calculate how many grams of Ba3(PO4)2 there are in 0.00817 moles.

  18. Step 5. Convert moles of Ba3(PO4)2 to grams: We now need g of Ba3(PO4)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 x 601.9 g = 4.92 g. 1 mol

  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 = 65.39 g/mol AgNO3 = 107.9 + 14 + (3 x 16) = 169.9 g/mol Zn = 2.0 g x 1 mol = 0.0305 mol 65.39 g AgNO3 = 2.50 g x 1 mol = 0.0147 mol 169.9 g

  21. Step 2. Guess limiting reagent Zn(s) + 2 AgNO3(aq)  Zn(NO3)2+ 2 Ag(s) 1 mol 2 mol 0.0305 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 using the AgNO3 to calculate a factor that can be used to multiply through the equation. Factor = 0.0147/2 = 0.00735 Zn(s) + 2 AgNO3(aq)  Zn(NO3)2+ 2 Ag(s) 1 mol 2 mol 0.00735 0.0147 We in fact have 0.0305 mol of Zn, which is more than the 0.00735 mol required, so AgNO3 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 yield x 100 Theoretical yield

  25. 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)

  26. 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

  27. 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.

  28. Calculation of theoretical yield: Convert0.0607 moles of BaSO4 to grams: Formula mass BaSO4 = 137.3 + 32.0 = 4 x 16.0 = 235.3 g/mol No of grams of BaSO4 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 x 100 % 14.29 g = 78.4 % yield

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