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STOICHIOMETRY Part II Mole-Mole Relationships

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Presentation on theme: "STOICHIOMETRY Part II Mole-Mole Relationships"— Presentation transcript:

1 STOICHIOMETRY Part II Mole-Mole Relationships
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3 Objectives Explain how balanced equations apply to both chemistry and everyday life. Interpret balanced chemical equations in terms of moles. Construct “mole ratios” from balanced chemical equations, and apply these ratios in mole-mole stoichiometric calculations. 3

4 Why Is This Important? Many things that we use are manufactured from chemicals. Cleaning products, cosmetics, medications, food products For a manufacturer to make a profit, the cost of making these items must be less than the money users pay for them. Chemical processes carried out in industry must be economical. This is where balanced equations can help

5 Application to Chemistry
Take the following chemical equation N H2  2NH3 The key is the “coefficient ratio” (aka the “mole ratio”)

6 The mole ratio is the critical step in all stoichiometry problems!
A mole ratio converts moles of one compound in a balanced chemical equation into moles of another compound 6

7 The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chemical reaction 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3 N2 and H2 will always react to form NH3 in this 1:3:2 ratio of moles If you started with 10 moles of N2 it would take 30 moles of H2 and would produce 20 moles of NH3

8 For Example 2 moles 1 mole 2 moles 2 : 1 : 2 2Mg + O2 → 2MgO
2 : : 1 mole 0.5 mole mole 4 moles 2 moles 4 moles 3.4 moles 1.7 moles moles etc etc etc 8

9 Let’s Practice 2 KClO3 → 2 KCl + 3 O2 2 moles 3 moles 2 : 3 6 moles
1) How many moles of KClO3 must decompose to produce 9 moles of oxygen gas? 2 KClO3 2 KCl + 3 O2 2 moles 3 moles 2 : 3 6 moles 9 moles 9

10 More practice 2) For the following reaction 2 SO2 + O2 ---> 2 SO3
If you have 4 moles of SO2, how many moles of O2 would you need? How many moles of SO3 would you produce?

11 More practice 3) For the following reaction: 2 H2 + O2 ---> 2 H2O
How many moles of H2O are produced when 5.00 moles of oxygen are used? If 3.00 moles of H2O are produced, how many moles of oxygen must be consumed? How many moles of hydrogen gas must be used, given the data in problem two?

12 More practice 4) For the following reaction:
Fe2O3 + 3 CO ---> 2 Fe + 3 CO2 How many moles of CO2 are produced if you start with 10 moles of Fe2O3 ? How many moles of CO would you want with 10 moles of Fe2O3 ? How many moles of Fe would be produced?

13 5) The reaction below shows the synthesis of aluminum oxide
Let’s make this harder! 5) The reaction below shows the synthesis of aluminum oxide 3O2 + 4Al → 2Al2O3 If you only had 18 moles of Al, how much product could you make? If you wanted to produce 2.4 moles of product, how many moles of each reactant would you need?

14 Solving #1 3O2 + 4Al → 2Al2O3 Known: 18 mol Al Unknown: ? mol Al2O3
18 mol Al x 2 mol Al2O3 = 9.0 mol Al2O3 4 mol Al Mole Ratio

15 Solving #2 3O2 + 4Al → 2Al2O3 Known: 2.4 mol Al2O3
Unknown: ? mol Al ? Mol O2 3O2 + 4Al → 2Al2O3 4 mol Al 2.4 mol Al2O3 x = 4.8 mol Al 2 mol Al2O3 3 mol O2 2.4 mol Al2O3 x = 3.6 mol O2 2 mol Al2O3

16 Work on your own To solve the problems on the
Mole to Mole Ratio worksheet

17 STOICHIOMETRY – Part II continues Mass-Mass Relationships
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18 Objectives Take mole-mole stoichiometric calculations to the next level. Interpret balanced chemical equations in terms of mass. Since we don’t have a mole scale, figure out how much of a product we would produce from a certain amount of reactant! 18

19 Reaction for the synthesis of aluminum oxide
Recall Our Example Reaction for the synthesis of aluminum oxide 3O2 + 4Al → 2Al2O3 If you only had 18 moles of Al, how many moles of product could you make? 2 mol Al2O3 18 mol Al x = 9.0 mol Al2O3 4 mol Al 19

20 Let’s Take This One Step Further …
How can would we go about figuring out… How many grams of Al2O3 would that be? 20 20

21 How to Relate Grams and Moles?
The molar mass of an element or a compound is the mass of one mole of that substance. Molar mass is the key to converting between grams and moles. molar mass Grams Moles 21 21

22 Molar Mass of Al2O3 = 2(26.98) + 3(16.00) =101.96 g/mol 9 moles x
grams = grams 1 1 mole 22 22

23 Why Calculate With Mass?
No lab balance measures moles directly; generally mass is the unit of choice. From the mass of 1 reactant or product, the mass of any other reactant or product in a given chemical equation can be calculated, provided you have a balanced equation. As in mole-mole calculations, the unknown can be either a reactant or a product. 23

24 General Stoichiometry Steps
Write a balanced equation. 2. Identify your known & unknown. 3. Set up your conversion table. 4. Make sure answer makes sense. 24 24

25 Examples CaC2 + H2O →C2H2 + Ca(OH)2
1) Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2) according to this equation: CaC2 + H2O →C2H2 + Ca(OH)2 How many grams of C2H2 are produced by adding water to 5.0 grams of CaC2? 25

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27 Problem #1 grams 5.0 grams ? _CaC2+ _H2O →_C2H2 + _Ca(OH)2 mole moles

28 BaCl2 + Na3PO4  Ba3(PO4)2 + NaCl
Examples 2) What mass of BaCl2 is needed to react completely with 46.8 grams Na3PO4 according to the equation below? BaCl2 + Na3PO4  Ba3(PO4)2 + NaCl 28

29 Problem #2 _BaCl2 + _Na3PO4  _Ba3(PO4)2 + _NaCl ? grams 46.8 grams
mole moles

30 Example #3 FeS + HCl  H2S + FeCl2
3) Use the equation below to determine what mass of FeS must react to form 326 grams of FeCl2. FeS + HCl  H2S + FeCl2 30

31 Problem #3 grams ? 326 grams _FeS + _HCl  _H2S + _FeCl2 mole moles

32 Why Is This Important? We can use this to determine how much of any reactant we have to add to achieve a certain amount of a particular product. We can use this to determine how much of any product we should get when a certain amount of a particular reactant is used. We can use variations of these to determine % yield or the limiting reagent in a reaction. We won’t be doing this now – but we will briefly! 32 32

33 Stop – Work on your own Work on your own on:
Mass to Mass Stoichiometry problems


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