Perform calculations to determine the percentage yield percentage yield of a reactionPerform calculations to determine the Atom Economy Atom Economy of a reaction

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2.2.8 Percentage yieldPercentage yieldAtom EconomyAtom Economy

Calculating Percentage (%) YieldCalculating Percentage (%) Yield2.3g of sodium reacts with an excess of chlorine to produce 4.0g of sodium chloride.

(Ar reactants: Na=23 Cl=35.5 Mr 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) + Cl2(g) 2NaCl(s)

2.3g Na = 2.3 mol Na23

= 0.1 mol Na

Theoretically 0.1 mol Na should yield 0.1 mol NaCl

Calculating Percentage (%) YieldCalculating Percentage (%) YieldIf 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) + O2(g) 2MgO(s)

(Ar reactants: Mg=24 O=16 Mr product: MgO= 40)

1.2g Mg = 1.2 mol Mg24

= 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

Calculating Percentage (%) YieldCalculating Percentage (%) YieldIf 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)+ CaCO3(s) H2O(l) + CO2(g) + CaCl2(s)

(Mr values are: CaCO3 = 100 CaCl2 = 111)

2g CaCO3 = 2 mol CaCO3100

= 0.02 mol CaCO3

Theoretically 0.02 mol CaCO3 should yield 0.02 mol CaCl2

111 x 0.02 = Theoretical Yield of CaCl2 = 2.22g

% Yield = Actual yield x 100% Theoretical yield

2.2 8 Percentage Yield

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Questions 1 and 2

Key Definition : A Limiting Reagent is . .

CH3CH(Cl) CH3 + NaOH → CH3CH(OH) CH3 + NaCl

3.295 g 2.955 g

Molar mass78.5 g 60 g

Moles3.925 / 78.5 2.955 / 60

0.05 0.04925

So moles we should have got is = 0.05or 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

C2H5OH + CH3COOH → CH3COOC2H5 + H2O

4.0 g 4.5 g 5.5 g

Molar mass46 g 60 g 88 g

Moles4.0 / 46 g 4.5 / 60 g 5.5 / 88 g

0.087 0.075 0.0625

So moles we should have got is = 0.075or 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

Atom EconomyA sample of magnetite iron ore contains 76% of the iron oxide compound Fe3O4 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:

Fe3O4 + 2C ==> 3Fe + 2CO2

(b) What is the atom economy of the carbon

reduction reaction?

(c) Will the atom economy be smaller, the same, or greater, if the reduction involves carbon monoxide (CO) rather than carbon (C)? explain?

Fe3O4 + 2C ==> 3Fe + 2CO2

Fe3O4 + 4CO ==> 3Fe + 4CO2