1. percentage yield Perform calculations to determine the percentage yield of a reaction Atom Economy Perform calculations to determine the Atom Economy of a reaction Page 162-163 Percentage yield 2.2.8 Percentage yield Atom Economy

2. Calculating Percentage (%) Yield 2.3g of sodium reacts with an excess of chlorine to produce 4.0g of sodium chloride. (A r reactants: Na=23 Cl=35.5 M r product: NaCl= 58.5) 58.5 x 0.1 =Theoretical yield of NaCl =5.85g What is the percentage yield? % Yield = Actual yield x 100% Theoretical yield % Yield = 4.0g x 100% = 5.85g 68% 2Na (s) + Cl 2(g)  2NaCl (s) 2.3g Na = 2.3 mol Na 23 = 0.1 mol Na Theoretically 0.1 mol Na should yield 0.1 mol NaCl

3. Calculating Percentage (%) Yield If 1.2g of magnesium reacts with an excess of oxygen to produce 0.8g of magnesium oxide… What is the percentage yield? % Yield = 0.8g x 100% = 2g 40% 2Mg (s) + O 2(g)  2MgO (s) (A r reactants: Mg=24 O=16 M r product: MgO= 40) 1.2g Mg = 1.2 mol Mg 24 = 0.05 mol Mg Theoretically 0.05 mol Mg should yield 0.05 mol MgO 40 x 0.05 =Theoretical yield of MgO =2g % Yield = Actual yield x 100% Theoretical yield

4. Calculating Percentage (%) Yield If 2g of calcium carbonate reacts with an excess of hydrochloric acid to produce 1.11 g of calcium chloride…. What is the percentage yield? % Yield = 1.11 x 100 = 2.22 50% 2HCl (aq) + CaCO 3(s)  H 2 O (l) + CO 2(g) + CaCl 2(s) (M r values are: CaCO 3 = 100 CaCl 2 = 111) 2g CaCO 3 = 2 mol CaCO 3 100 = 0.02 mol CaCO 3 Theoretically 0.02 mol CaCO 3 should yield 0.02 mol CaCl 2 111 x 0.02 =Theoretical Yield of CaCl 2 =2.22g % Yield = Actual yield x 100% Theoretical yield

5. 2.2 8 Percentage Yield Page 162-163 Questions 1 and 2 Key Definition : A Limiting Reagent is..

6. CH 3 CH(Cl) CH 3 + NaOH → CH 3 CH(OH) CH 3 + NaCl 3.295 g 2.955 g Molar mass 78.5 g 60 g Moles 3.925 / 78.5 2.955 / 60 0.05 0.04925 So moles we should have got is = 0.05 or mass should have been = 0.05 x 60 = 3.00 g so yield is 0.04925/0.050 x100 = 98.5 % or using grams 2.955/3.00 x 100 = 98.5 % Question 1

7. C 2 H 5 OH + CH 3 COOH → CH 3 COOC 2 H 5 + H 2 O 4.0 g 4.5 g 5.5 g Molar mass 46 g 60 g 88 g Moles 4.0 / 46 g 4.5 / 60 g 5.5 / 88 g 0.087 0.075 0.0625 So moles we should have got is = 0.075 or mass should have been = 0.075 x 88 = 6.6g so yield is 0.0625/0.075 x 100 = 83.3 % or using grams 5.5/6.6 x100 = 83.3 % Question 2

10. Atom Economy A sample of magnetite iron ore contains 76% of the iron oxide compound Fe 3 O 4 and 24% of waste silicate minerals. (a) What is the maximum theoretical mass of iron that can be extracted from each tonne (1000 kg) of magnetite ore by carbon reduction? [ Atomic masses: Fe = 55.8, C = 12 and O = 16 ] o The reduction equation is: Fe 3 O 4 + 2C ==> 3Fe + 2CO 2 (b) What is the atom economy of the carbon reduction reaction?

11. (c) Will the atom economy be smaller, the same, or greater, if the reduction involves carbon monoxide (CO) rather than carbon (C)? explain? Fe 3 O 4 + 2C ==> 3Fe + 2CO 2 Fe 3 O 4 + 4CO ==> 3Fe + 4CO 2