Calculations From Chemical Equations

Transcription

1 Calculations From Chemical Equations Chapter 9 Hein and Arena Version 1.1 Eugene Passer Chemistry Department Bronx Community 1 College John Wiley and Sons, Inc.

2 A Short Review 2

3 The molar mass of an element is its atomic mass in grams/mol. It contains x atoms (Avogadro s number) of the element. The molar mass of a compound is the sum of the atomic masses of all its atoms of each element. 3

4 Calculate the molar mass of C 2 H 6 O. 2 C = 2(12.01 g) = g 6 H = 6(1.01 g) = 6.06 g 1 O = 1(16.00 g) = g g/mol 4

5 For calculations of mole-mass-volume relationships in a chemical equation. The chemical equation must be balanced by using smallest whole number coefficients. The coefficient in front of a formula represents the number of moles, atoms, ions or molecules, of the reactant or product. The equation is balanced. 2Al + Fe 2 2 O 3 Fe + Al 2 O 3 2 mol 1 mol 2 mol 1 mol 5

6 Introduction to Stoichiometry: The Mole Ratio Method 6

7 Stoichiometry: The area of chemistry that deals with the quantitative relationships between reactants and or products. Mole Ratio: A ratio between the moles of any two substances involved in a chemical reaction (Conversion Factor). The coefficients used in mole ratio expressions are derived from the coefficients used in the balanced equation. 7

8 Examples 8

9 N + 3H 2NH mol 3 mol 2 mol 1 mol N 2 3 molh 2 9

10 N + 3H 2NH mol 3 mol 2 mol 3 molh 2 2 mol NH 3 10

11 The mole ratio is used to convert moles of one substance to moles of another substance in a stoichiometry problem. mole ratio mol. H 2 mol. N 2 The mole ratio is used to solve every type of stoichiometry problem. 11

12 The Mole Ratio Method: Step by Step 1. Convert the quantity of starting substance to moles (if it is not already in moles) 2. Convert the moles of starting substance to moles of desired substance. 3. Convert the moles of desired substance to the units specified (i.e. grams, molecules, atoms, etc.) in the problem. 12

13 Mole-Mole Calculations 13

14 Calculate the number of moles of phosphoric acid (H 3 PO 4 ) formed by the reaction of 10 moles of sulfuric acid (H 2 SO 4 ). Ca 5 (PO 4 ) 3 F + 5H 2 SO 4 3H 3 PO 4 + HF + 5CaSO 4 1 mol 5 mol 3 mol 1 mol 5 mol Step 1 Moles starting substance: 10.0 mol H 2 SO 4 Step 2 The conversion sequence is moles H 2 SO 4 moles H 3 PO 4 3 mol H PO mol H2SO4 x = 5 mol H2SO4 Mole Ratio 6 mol H3PO4 14

15 Calculate the number of moles of sulfuric acid (H 2 SO 4 ) that react when 10 moles of Ca 5 (PO 4 ) 3 F react. Ca 5 (PO 4 ) 3 F + 5H 2 SO 4 3H 3 PO 4 + HF + 5CaSO 4 1 mol 5 mol 3 mol 1 mol 5 mol Step 1 The starting substance is 10.0 mol Ca 5 (PO 4 ) 3 F Step 2 The conversion sequence is moles Ca 5 (PO 4 ) 3 F moles H 2 SO 4 Mole Ratio 5 mol H SO mol Ca 5(PO 4) 3F x = 1 mol Ca 5(PO 4) 3F 50 mol H2SO4 15

16 In the following reaction how many moles of PbCl 2 are formed if moles of NaCl react? 2NaCl(aq) + Pb(NO 3 ) 2 (aq) PbCl 2 (s) + 2NaNO 3 (aq) The conversion sequence is moles NaCl moles PbCl 2 1 mol PbCl moles of PbCl 2= moles NaCl mol PbCl2 2 mol NaCl 16

17 Mole-Mass Calculations 17

18 Examples 18

19 Calculate the number of moles of H 2 SO 4 necessary to yield 784 g of H 3 PO 4 Ca 5 (PO 4 ) 3 F+ 5H 2 SO 4 3H 3 PO 4 + HF + 5CaSO 4 The conversion sequence is grams H 3 PO 4 moles H 3 PO 4 moles H 2 SO 4 Mole Ratio 1 mol H3PO4 784 g H3PO g H PO mol H2SO 4 = 3mol H PO mol H2SO4 19

20 Calculate the number of grams of H 2 required to form 12.0 moles of NH 3. N 2 + 3H 2 2NH 3 The conversion sequence is moles NH 3 moles H 2 grams H 2 Mole Ratio 3 mol H mol NH3 2 mol NH 1mol H g H = 36.4 g H 2 20

21 Mass-Mass Calculations 21

22 Calculate the number of grams of NH 3 formed by the reaction of 112 grams of H 2. N 2 + 3H 2 2NH 3 grams H 2 moles H 2 moles NH 3 grams NH 3 1 mol H g H g H2 3 mol H2 2 mol NH 17.0 g NH 3 = 1 mol NH g NH 3 22

23 Limiting-Reactant and Yield Calculations 23

24 Limiting Reactant 24

25 A limiting reactant is a reactant in a chemical reaction with insufficient mass to react with all the mass of the other reactant(s) present. The limiting reactant limits the amount of product that can be formed. 25

26 How many From bicycles From eight three wheels From pedal four assemblies frames four can be assembled bikes three can be bikes constructed. bikes can can be constructed. be from the parts shown? The limiting part is the number of pedal assemblies

27 Steps Used to Determine the Limiting Reactant 27

28 1. Calculate the moles of each reactant (This is what you HAVE). 2. Convert moles of each reactant to moles of the other reactant (This is what you NEED). 3. Determine limiting reactant by: HAVE NEED > 0 ; NOT LIMITING HAVE NEED < 0 ; LIMITING 4. Excess reactant is the difference between: HAVE NEED of NON- LIMITING reactant. 28

29 Examples 29

30 How many grams of H 2 will be produced if 15.0 g of HCl are reacted with 7.5 g Mg? Assume 100% yield. Mg + 2HCl MgCl 2 + H 2 Step 1: Step 2: Step 3: Step 4: Calculate moles of reactants you have. Calculate moles of reactants you need. Determine limiting reactant. Calculate desired quantity based on limiting reactant. 30

31 How many grams of H 2 will be produced if 15.0 g of HCl are reacted with 7.5 g Mg? Assume 100% yield. Mg + 2HCl MgCl 2 + H 2 Have Need Moles have (Step 1): x mol Mg = 7.5g Mg x 1mol/24.31g Mg = 0.31mol. Mg x mol HCl = 15.0g HCl x 1mol/36.45g HCL = 0.41 mol HCl 31

32 How many grams of H 2 will be produced if 15.0 g of HCl are reacted with 7.5 g Mg? Assume 100% yield. Mg + 2HCl MgCl 2 + H 2 Have Need Moles need (Step 2): x mol Mg = 0.41mol HCl x 1m Mg/2 m HCl = 0.21mol. Mg x mol HCl = 0.31 mol Mg x 2 m HCl/1m Mg = 0.62 mol HCl 32

33 How many grams of H 2 will be produced if 15.0 g of HCl are reacted with 7.5 g Mg? Assume 100% yield. Mg + 2HCl MgCl 2 + H 2 Have Need Limiting reactant (Step 3): The limiting reactant is HCl, since we have less than we need: have - need < 0; =

34 How many grams of H 2 will be produced if 15.0 g of HCl are reacted with 7.5 g Mg? Assume 100% yield. Mg + 2HCl MgCl 2 + H 2 The grams of H 2 is calculated from the mass of the limiting reactant HCl. g H g HCl x 1mol. HCl x 1mol. H 2 x 2.02g H 2 = 36.45g HCl 2 mo HCl 1 mol H g H 2 34

35 Suppose grams of carbon and grams of steam react. How many grams of hydrogen will be produced? Assume 100% yield. C (s) + H 2 O (g) CO (g) + H 2 (g)

36 C (s) + H 2 O (g) CO (g) + H 2 (g) Moles have Moles need moles xs H 2 O Therefore, the carbon is the limiting reactant, since we have less carbon than we need to completely react with all the water we have. The mass of hydrogen is calculated using the limiting reactant; g C x 1 mol C x 1 mol H 2 x 2.02 g H 2 = g H g C 1 mol C 1 mol H 2

37 Reaction Yield 37

38 The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation. 38

39 Many reactions fail to give a 100% yield of product. This occurs because of: side reactions purification steps reversible reactions 39

40 The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant. The actual yield is the amount of product actually obtained from a given amount of reactant. 40

41 The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100. actual yield theoretical yield x 100 = percent yield 41

42 Silver bromide was prepared by reacting g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if g of silver bromide was obtained from the reaction: MgBr 2 (aq) + 2AgNO 3 (aq) 2AgBr(s) + Mg(NO 3 ) 2 (aq) Step 1 Determine the theoretical yield by calculating the grams of AgBr that can be formed. The conversion sequence is g MgBr 2 mol MgBr 2 mol AgBr g AgBr 1 mol MgBr 2 2 mol AgBr g AgBr g MgBr2 1 mol AgBr g AgBr g MgBr2 1 mol MgBr2 42

43 Silver bromide was prepared by reacting g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if g of silver bromide was obtained from the reaction: MgBr 2 (aq) + 2AgNO 3 (aq) 2AgBr(s) + Mg(NO 3 ) 2 (aq) Step 2 Calculate the percent yield. must have same units actual yield percent yield = x 100 theoretical yield must have same units percent yield = g AgBr x 100 = g AgBr 91.9% 43

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