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Stoichiometry. The Arithmetic of Equations 12.1.1 – I can explain how balanced equations apply to both chemistry and everyday life. 12.1.2 – I can interpret.

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Presentation on theme: "Stoichiometry. The Arithmetic of Equations 12.1.1 – I can explain how balanced equations apply to both chemistry and everyday life. 12.1.2 – I can interpret."— Presentation transcript:

1 Stoichiometry

2 The Arithmetic of Equations

3 – I can explain how balanced equations apply to both chemistry and everyday life – I can interpret balanced chemical equations in terms of moles, representative particles, mass, and gas volume at STP – I can identify the quantities that are always conserved in chemical reactions.

4  A balanced chemical equation provides the same kind of quantitative information that a recipe does.

5  Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction.

6  Tiny tike has decided to make 288 tricycles each day. How many tricycle seats, wheels, and pedals are needed?

7  Stoichiometry – the calculation of quantities in chemical reactions.

8  A balanced chemical equation can be interpreted in terms of different quantities, including; numbers of atoms, molecules, or moles, mass, and volume.

9  A balanced equation indicates that the number and type of each atom that makes up each reactant also makes up each product.

10  Here the coefficients tell you how many molecules will react and form.  Much like how many atoms.

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12  A balanced equation also tells you the number of moles of reactants and products.  The coefficients tell you this.  You’ll use this most often.

13  A balanced equation obeys the law of conservation of mass.  Using a mole relationship you can relate number of moles to mass of either reactants or products.

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15  If you are at STP the equation tells you about volumes of gases.  You can use mole relationships for this also.

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17  Mass and atoms are conserved in every chemical reaction.

18  Balance the following equation. ___C 2 H 4 (g) + ___O 2 (g) → ___CO 2 (g) + ___H 2 O(g)  Interpret the balanced equation in terms of relative numbers of moles, volumes of gas at STP, and masses of reactants and products.

19 Chemical Calculations

20 – I can construct mole ratios from balanced chemical equations and apply these ratios in stoichiometric calculations – I can calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles, and volumes of gases at STP.

21  Mole ratio – a conversion factor derived from the coefficients of a balanced chemical reaction interpreted in terms of moles.

22  In chemical calculations, mole ratios are used to convert between moles of reactant and moles of product, between moles of products, or between moles reactants.

23  4Al(s) + 3O 2 (g) → 2Al 2 O 3 (s)  Write six mole ratios that can be derived from this equation.

24  The easiest way to see these is to do an example.  W is the unknown, G is the given quantity.  a and b are the coefficients from the balanced chemical equation.

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26  4Al(s) + 3O 2 (g) → 2Al 2 O 3 (s)  How many moles of aluminum are needed to form 3.7 moles of aluminum oxide?

27  In a reaction things are not measured in moles (no scale does this).  Instead things are measured in mass then transferred to moles.

28 1. Change the mass of G to moles of G (mass G → mol G) by using the molar mass of G.

29 2. Change the moles of G to moles of W (mol G →mol W) by using the mole ratio from the balanced equation.

30 3. Change the moles of W to grams of W (mol W → mass W) by using the molar mass of W.

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32  Acetylene gas (C 2 H 2 ) is produced by adding water to calcium carbide (CaC 2 ). CaC 2 (s) + 2H 2 O(l) → C 2 H 2 (g) + Ca(OH) 2 (aq)  How many grams of acetylene are produced by adding water to 5.00g of calcium carbide?

33  In a typical stoichiometric problem, the given quantity is first converted to moles.  Then the mole ratio from the balanced equation is used to calculate the number of moles of the wanted substance.

34  Finally, the moles are converted to any other unit of measure related to the unit mole, as the problem requires.

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36  How many molecules of oxygen are produced by the decomposition of 6.54g of potassium chlorate (KClO 3 )? 2KClO 3 (s) → 2KCl(s) + 3O 2 (g)

37  The equation for the combustion of carbon monoxide is: 2CO(g) + O 2 → 2CO 2 (g)  How many liters of oxygen are required to burn 3.86L of carbon monoxide?

38 Limiting Reagent and Percent Yield

39 – I can identify the limiting reagent in a reaction – I can calculate theoretical yield, actual yield, or percent yield given appropriate information.

40  When cooking you know you need the right amounts of ingredients for the recipe to turn out.  In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.

41  Limiting reagent – determines the amount of product that can be formed in the reaction.  This is the one used up first in a reaction.  The reaction can only “go” until this reactant is completely used up.  Excess reagent – reactant that is not completely used up in a reaction.

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43 How would the amount of products formed if you started with four molecules of N 2 and three molecules H 2 ?

44  The equation for the complete combustion of ethene (C 2 H 4 ) is: C 2 H 4 (g) + 3O 2 (g) → 2CO 2 (g) + 2H 2 O(l)  If 2.70mol of C 2 H 4 is reacted with 6.30mol of O 2, identify the limiting reagent.

45  The reactant that is present in the smaller amount by mass or volume is not necessarily the limiting reagent.

46  The heat from an acetylene torch is produced by burning acetylene (C 2 H 2 ) in oxygen: 2C 2 H 2 + 5O 2 → 4CO 2 + 2H 2 O  How many grams of water can be produced by the reaction of 2.4 mol C 2 H 2 with 7.4 mol O 2 ?

47  Your grades are usually expressed as a percent  Chemists use similar calculations when products are formed based on balanced equations.  In theory all reactions would produce at 100%.  In reality they don’t.

48  Theoretical yield – the maximum amount of product that could be formed from given amounts of reactants.  Actual yield – the amount of product that actually forms when the reaction is carried out.

49  Percent yield – ratio of the actual yield to the theoretical yield expressed as a percent.

50  The percent yield is a measure of the efficiency of a reaction carried out in the laboratory.  Percent yield can be lowered by:  Impure reactants.  Loss of product in filtration or transferring.  If reactants or products have not been carefully measured.

51  When 84.8g of iron (III) oxide reacts with an excess of carbon monoxide, iron is produced. Fe 2 O 3 (s) + 3CO → 2Fe(s) + 3CO 2 (g)  What is the theoretical yield of iron?

52  If 50.0g of silicon dioxide is heated with an excess of carbon, 27.9g of silicon carbide is produced. SiO 2 (s) + 3C(s) → SiC(s) + 2CO(g)  What is the percent yield of this reaction?


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