Percent Yield = Actual yield of product x 100% Theoretical yield of product

Transcription

1 Yields of Chemical Reactions In the stoichiometry examples so far we have made the unstated assumption that the reaction goes to completion, that all reactants are converted to products. In actual fact, the amount of product actually formed in a real reaction, called actual yield, is less than the amount that calculations predict ( theoretical yield). Reaction yields are expressed as percent Yield. Percent Yield = Actual yield of product x 100% Theoretical yield of product Example Diethyl ether (C 4 H 10 O) is prepared by treatment of ethyl alcohol (C 2 H 6 O) with an acid catalyst. What is the percent yield of the reaction if 28.0 g of ethyl ether is obtained from the reaction of 40.0 g of ethyl alcohol? The balanced chemical equation is:. 2 C 2 H 6 O (l) acid C 4 H 10 O (l) + H 2 O (l) Solution We need to calculate the amount of C 4 H 10 O that could be theoretically produced from 40.0 g of C 2 H 6 O and then compare that to the actual yield (28.0 g) 1. Calculate molar masses of reactant and product molar mass of C2H6O = (2 x 12.01) + (6 x 1.01) = g/ mol molar mass of C4H10O = (4 x 12.01) + (10 x 1.01) = g/mol 2. Convert mass of C 2 H 6 O to mol C 2 H 6 O 40.0 g of C 2 H 6 O x 1 mol of C 2 H 6 O = mol C 2 H 6 O g C 2 H 6 O 3. Calculate theoretical yield of C 4 H 10 O in mol. Use reaction equation coefficients mol C 2 H 6 O x 1 mol C 4 H 10 O = mol C 4 H 10 O 2 mol C 2 H 6 O 4. Calculate the theoretical yield of C 4 H 10 O in mass mol C 4 H 10 O x g C 4 H 10 O = or g C 4 H 10 O 1 mol C 4 H 10 O 5. Calculate % yield % Yield = Actual yield x 100 = 28.0 g C 4 H 10 O x 100 = 87.0 % yield Theoretical yield 32.2 g C 4 H 10 O 1

2 Reactions with Limiting Amounts of Reactant Many reactions are carried out using an excess of one reactant (more than is needed According to stoichiometry). The other reactant will be present in limiting amounts and the extent to which a chemical reaction takes place depends on theamount of this reactant (called limiting reactant) For example: C 2 H 4 O + H 2 O C 2 H 6 O 2 (Balanced equation) ethylene oxide ethylene glycol If there are 3 molecules of C 2 H 4 O and 5 molecules of H 2 O then H 2 O is in excess and C 2 H 4 O is the limiting reactant Why? Because the coefficients of the reaction says you only need 3 molecules of water to react with 3 molecules of C 2 H 4 O. Two molecules of water are left in excess. To determine the LIMITING REACTANT To determine AMOUNT OF PRODUCT formed and To calculate the amount of EXCESS REACTANT REMAINING 1. Calculate how many moles of each reactant are present and mole ratio added. 2. Look at coefficients in the balanced equation to calculate the required mole ratio of the two reactants. Compare with mole ratio in step 1. Identify limiting and excess reactants. 3. To determine the amount of product formed, use the number of moles of limiting reactant and the reaction coefficients to calculate theoretical yield of product. Convert to mass. 4. Subtract the number of moles of excess reactant consumed from moles of excess reactant initially present. Convert moles to grams. This is mass of reactant in excess. 2

3 EXAMPLE: Boron sulfide, B 2 S 3 reacts violently with water to form dissolved boric acid H 3 BO 3 and hydrogen sulfide gas. B 2 S 3 (s) + 6 H 2 O (l) 2 H 3 BO 3 (aq) + 3 H 2 S (g) If 5.88 g of B 2 S 3 is mixed with 7.85 g of water. a) Which reactant is limiting and which is in excess?. b) How many grams of boric acid product are theoretically produced? c) How many grams of the excess reactant remain? SOLUTION: a) Determine mole ratio of the reactants 5.88 g B 2 S 3 x 1 mol B 2 S 3 = mol B 2 S g B 2 S g H 2 O x 1 mol H 2 O = mol H 2 O g H 2 O mol of H 2 O = mol = 8.74 (mol ratio of reactants added) mol of B 2 S mol the required mole ratio (from the coefficients of balanced equation) is: Mol of H 2 O = 6 mol H 2 O = 6 (mole ratio required) Mol of B 2 S 3 1 mol B 2 S 3 Compared to ratio of 8.74 ( previous slide) there is more H 2 O than required Therefore H 2 O is reactant in excess and B 2 S 3 is the limiting reactant b) Theoretical yield of H 3 BO 3 product (based on moles of limiting reactant!) mol B 2 S 3 x 2 mol H 3 BO 3 x g H 3 BO 3 = 6.17 g H 3 BO 3 1 mol B 2 S 3 1 mol H 2 BO 3 c) 1st calculate quantity of excess H 2 O consumed then H 2 O which is not consumed? mol of B 2 S 3 consumed = mol B 2 S 3 mol of H 2 O consumed = mol B 2 S 3 x 6 mol H 2 O = mol H 2 O consumed 1 mol B 2 S H 2 O not consumed = mol H 2 O (initial) mol H 2 O (consumed) = mol mass of H 2 O not consumed = mol x g H 2 O = 2.47 g H 2 O not consumed 1 mol H 2 O 3

4 Concentrations of reactants in Solution: Molarity Most reactions are carried out in solution. Why? Because reactants must have considerable mobility for the reaction to occur. Solute- the substance dissolved in solution Concentration- describes the relative amount of solute to a solvent. Most common unit used is: Molarity (M) =the number of moles of solute dissolved in one liter of solution (units of mol/l) useful conversion factors: or Molarity (M) = moles of solute (mol) volume of solution (V) moles of solute = Molarity x Volume = M x V or Volume = moles of solute = mol Molarity M EXAMPLE 1: What is the molarity of a solution prepared by dissolving 16.8 g of NaOH and diluting to a final volume of 2.50 L? Mol of NaOH = 16.8 g NaOH x 1mol NaOH = mol NaOH g NaOH Molarity = mol of NaOH = mol NaOH = M NaOH solution volume 2.50 L EXAMPLE 2: How many moles of K 2 Cr 2 O 7 are present in ml of a M solution of K 2 Cr 2 O 7? Step 1. (Convert volume to L) ml x 1 L = L 1000 ml Step 2. moles = Molarity x V = mol/l x L = 6.25 x 10-3 mol K 2 Cr 2 O 7 4

5 EXAMPLE 3: A particular reaction requires that you use 36.7 g of H 2 SO 4. How many ml of a 5.00 M solution of H 2 SO 4 are required to provide 36.7 g? Step 1. Determine mol of H 2 SO 4 required: 36.7 g H 2 SO 4 x 1 mol H 2 SO 4 = mol g H 2 SO 4 Step 2. V = mol required = mol H 2 SO 4 Molarity 5.00 mol/l = L or 74.8 ml PROBLEM How many moles of ClO - 4 (aq) are in ml of M Mg (ClO 4 ) 2 Ans: mol ClO 4 - ions 5

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