Stoichiometry. mass relationships. how much reactant is needed to yield a certain amount of product

Transcription

1 Mass Relationships

2 Stoichiometry mass relationships how much reactant is needed to yield a certain amount of product

3 Amounts of Reactants and Products

4 The mole method 1. Write and balance the equation. 2. Convert the given quantities into moles. 3. Use the coefficients in the balanced equation to relate the number of moles of known substances to the desired unknown one. 4. Convert to desired units. 5. Check your answer.

5 Stoichiometry Molar ratio of Y to X n y n x Moles of X Moles of Y Mass of X X Y Mass of Y

6 Example How many grams of nitrogen dioxide can be formed by reaction of 1.44 g of nitrogen monoxide with oxygen? 2NO + O 2 2NO 2

7 Stoichiometry Molar ratio of Y to X n y n x Moles of X Moles of Y 1.44 g of NO Mass of NO 2 2NO + O 2 2NO 2

8 Stoichiometry Molar ratio of Y to X n y n x Moles NO Moles of Y 1.44 g of NO Mass of NO 2 2NO + O 2 2NO 2

9 Stoichiometry Molar ratio NO 2 to NO Moles NO Moles of Y 1.44 g of NO Mass of NO 2 2NO + O 2 2NO 2

10 Stoichiometry Molar ratio NO 2 to NO Moles NO Moles NO g of NO Mass of NO 2 2NO + O 2 2NO 2

11 Stoichiometry Molar ratio NO 2 to NO Moles NO Moles NO g of NO 2.21 g of NO 2 2NO + O 2 2NO 2

12 Example 2NO + O 2 2NO g NO x 1mol NO 30g NO x 2mol NO 2 2mol NO x 46g NO 2 1mol NO 2 = 2.21g NO 2

13 Stoichiometric Calculations Starting with the mass of Substance A, you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it s a product) or used (if it s a reactant). Stoichiometry 2012 Pearson Education, Inc.

14 Stoichiometric Calculations C 6 H 12 O O 2 6 CO H 2 O Starting with 1.00 g of C 6 H 12 O 6 we calculate the moles of C 6 H 12 O 6 use the coefficients to find the moles of H 2 O and then turn the moles of water to grams Pearson Education, Inc. Stoichiometry

15 Limiting Reagents

16 Limiting Reagent Reactants are not always present (or available) in stoichiometric quantities. One reactant may be present in quantities such that it is completely consumed while excess amounts of other reactants remain. - called limiting reactant or limiting reagent The limiting reagent will limit the amount of product produced.

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18 Limiting Reactants The limiting reactant is the reactant present in the smallest stoichiometric amount. In other words, it s the reactant you ll run out of first (in this case, the H 2 ) Pearson Education, Inc. Stoichiometry

19 Limiting Reactants In the example below, the O 2 would be the excess reagent Pearson Education, Inc. Stoichiometry

20 Example How many moles of MgCl 2 will be produced? Mg + Cl 2 MgCl 2 Start 1 mol 1 mol 0 Finish mol

21 Example How many moles of MgCl 2 will be produced? Mg + Cl 2 MgCl 2 Start 1 mol 2 mol 0 Finish 0 1 mol 1 mol magnesium is the limiting reagent 1 mol of chlorine will be left unchanged

22 Limiting Reagent Molar ratio Y to X n y n x Moles of X Moles of W Moles of Y Mass of X Mass of W W + X Y Mass of Y

23 Compare molar ratio W to X to their coefficients in balancd equation; identify LR Molar ratio Y to LR Moles of X Moles of W Moles of Y Mass of X Mass of W W + X Y Mass of Y

24 Example Determine the limiting reagent and the amount of PI 3 produced when 6.00g P 4 reacts with 25.0g of I 2. P 4 + 6I 2 4PI 3 1mol P 6.00g P 4 x 4 =.0484mol P 4 124g P 4 1mol I 25.0g I 2 x 2 =.0984mol I 2 254g I 2

25 Example cont... Determine how much I 2 would be needed to react completely with the available amount of P 4. P 4 + 6I 2 4PI mol P 4 x 6mol I 2 1mol P 4 but only.0984mol I 2 is available I 2 is the limiting reagent = 0.290mol I 2 amount of I 2 needed

26 Example cont... the amount of PI 3 produced from the limiting reagent mol I 2 x 4mol PI 3 6mol I 2 x 412g PI 3 1mol PI 3 = 27.0g PI 3

27 another method Determine the limiting reagent and the amount of PI 3 produced when 6.00g P 4 reacts with 25.0g of I 2. P 4 + 6I 2 4PI g P 4 x 1mol P 4 124g P 4 =.0484mol P g I 2 x 1mol I 2 254g I 2 =.0984mol I 2

28 another method P 4 + 6I 2 4PI mol I 2 = mol P 4 The actual I 2 /P 4 ratio is less than the stiochiometic ratio 6 mol I 2 = 6 So there is not enough I2 1 mol P to 4 I 2 is the limiting reagent react with all the P 4

29 Example How much aluminum oxide is formed from 124 g of Al and 601 g Fe 2 O 3? 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 124 g 601 g 4.6 mol 3.8 mol excess of 1.5 mol limiting reactant

30 Example How much aluminum oxide is formed from 124 g of Al and 601 g Fe 2 O 3? 2Al + Fe 2 O 3 Al 2 O 3 + 2Fe 124 g 4.6 mol 2.3 mol = 234 g limiting reactant

31 Example From the reaction between of 10.0g of Hg and 9.0g of Br 2. What mass of which reagent is left unreacted? Hg + Br 2 HgBr g Hg x 1 molhg 200.6gHg = 4.99 x 10-2 molhg 9.0g Br 2 x 1 molbr 2 = 5.63 x 10-2 molbr gBr 2

32 Hg + Br 2 HgBr x 10-2 molhg Hg is limiting 1 molbr 2 x = 4.99 x 10-2 molbr 2 1 molhg moles of Br 2 needed to use up Hg available 159.8gBr 4.99 x 10-2 molbr 2 x 2 1 molbr 2 = 7.97 g Br 2 grams of Br 2 used 9.0g Br g Br 2 = 1.03 g Br 2 excess

33 Reaction Yield

34 Theoretical yield the amount of product that would result if all the limiting reagent reacted Actual yield The calculated yield the amount of product actually obtained from the reaction Almost always less than the theoretical yield

35 Percent Yield Actual yield %yield = x 100% Theoretical yield Determines how efficient a reaction is

36 Example In a certain industrial operation 3.54 x 10 7 g of TiCl 4 is reacted with 1.13 x 10 7 g of Mg. (a) Calculate the theoretical yield of Ti in grams. (b) Calculate the percent yield if 7.91 x 10 6 g are actually obtained. TiCl 4 + 2Mg Ti + 2MgCl 2

37 Calculate theoretical yield 3.54 x 10 7 g TiCl 4 x 1mol TiCl g TiCl 4 = 1.87 x 10 5 mol TiCl g Mg x 1mol Mg 24.31g Mg = 4.65 x 10 5 mol Mg 2mol Mg 1.87 x 10 5 mol TiCl 4 x 1mol TiCl 4 = 3.74 x 10 5 mol Mg there is more than enough Mg TiCl 4 is limiting

38 3.54 x 10 7 g TiCl 4 x 1mol TiCl 4 x 1mol Ti 187.7g TiCl 4 1mol TiCl 4 x 47.88g Ti 1mol Ti = 7.91 x 10 6 g Ti 100% x = 8.93 x 10 6 g Ti 8.93 x 10 6 g Ti Actual yield %yield = x 100% = 88.6% Theoretical yield

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