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USING EQUATIONS Nearly everything we use is manufactured from chemicals. › Soaps, shampoos, conditioners, cd’s, cosmetics, medications, and clothes. For a manufacturer to make a profit the cost of making any of these items can’t be more than the money paid for them. Chemical processes carried out in industry must be economical, this is where balanced equations help.

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USING EQUATIONS Equations are a chemist’s recipe. ◦ They tell chemists what amounts of reactants to mix and what amounts of products to expect. When you know the quantity of one substance in a reaction, you can calculate the quantity of any other substance consumed or created in the rxn. ◦ Quantity meaning the amount of a substance in grams, liters, molecules, or moles.

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USING EQUATIONS The calculation of quantities in chemical reactions is called stoichiometry.

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USING EQUATIONS Assume that the major components of a bike are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P). The finished bike has a “formula” of FSW 2 HP 2. The balanced equation for the production of 1 bike is. F +S+2W+H+2P FSW 2 HP 2

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USING EQUATIONS Now in a 5 day workweek, A company is scheduled to make 640 bikes. How many wheels should be in the plant on Monday morning to make these bikes? What do we know? ◦ Number of bikes = 640 bikes ◦ 1 FSW 2 HP 2 =2W (balanced eqn) What is unknown? ◦ # of wheels = ? wheels F +S+2W+H+2P FSW 2 HP 2

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The connection between wheels and bikes is 2 wheels per bike. We can use this information as a conversion factor to do the calculation. 640 FSW 2 HP 2 1 FSW 2 HP 2 2 W = 1280 wheels We can make the same kinds of connections from a chemical rxn eqn. N 2 (g) + 3H 2 (g) 2NH 3 (g) The key is the “coefficient ratio”.

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◦ The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chemical rxn. 1 mole of N 2 reacts with 3 moles of H 2 to produce 2 moles of NH 3. ◦ N 2 and H 2 will always react to form ammonia in this 1:3:2 ratio of moles. So if you started with 10 moles of N 2 it would take 30 moles of H 2 and would produce 20 moles of NH 3 N 2 (g) + 3H 2 (g) 2NH 3 (g)

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Using the coefficients, from the balanced equation to make connections between reactants and products, is the most important information that a rxn equation provides. › Using this information, you can calculate the amounts of the reactants involved and the amount of product you might expect. › Any calculation done with the next process is a theoretical number, the real world isn’t always perfect.

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MOLE – MOLE EXAMPLE The following rxn shows the synthesis of aluminum oxide. 3O 2 (g) + 4Al(s) 2Al 2 O 3 (s) If you only had 1.8 mols of Al how much product could you make? Given: 1.8 moles of Al Uknown: ____ moles of Al 2 O 3 3O 2 (g) + 4Al(s) 2Al 2 O 3 (s)

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MOLE – MOLE EXAMPLE Solve for the unknown: 1.8 mol Al 4 mol Al 2 mol Al 2 O 3 = 0.90mol Al 2 O 3 3O 2 (g) + 4Al(s) 2Al 2 O 3 (s) Mole Ratio

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MOLE – MOLE EXAMPLE 2 The following rxn shows the synthesis of aluminum oxide. 3O 2 (g) + 4Al(s) 2Al 2 O 3 (s) If you wanted to produce 24 mols of product how many mols of each reactant would you need? Given: 24 moles of Al 2 O 3 Uknown: ____ moles of Al ____ moles of O 2

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MOLE – MOLE EXAMPLE 2 Solve for the unknowns: 24 mol Al 2 O 3 2 mol Al 2 O 3 4 mol Al = 48 mol Al 3O 2 (g) + 4Al(s) 2Al 2 O 3 (s) 24 mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 = 36 mol O 2

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Learning check Sodium metal reacts with chlorine gas to produce sodium chloride. Write a balanced chemical equation. If 3.75 mol Na react with enough chlorine gas, how much sodium chloride is produced? (answer: 3.75 mol NaCl) CW p p375 #11,12

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MASS – MASS CALCULATIONS 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 reaction equation. As in mole-mole calculations, the unknown can be either a reactant or a product.

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MOLE A (mol) MASS A (g) MOLE B (mol) MASS B (g) Molar mass A g/mol (from P. table) Molar mass B g/mol (from P. table ) Mole ratio (from BALANCED equation)

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Acetylene gas (C 2 H 2 ) is produced by adding water to calcium carbide (CaC 2 ). How many grams of C 2 H 2 are produced by adding water to 5.00 g CaC 2 ? CaC 2 + 2H 2 O C 2 H 2 + Ca(OH) 2 MASS – MASS CALCULATIONS

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What do we know? Given mass = 5.0 g CaC 2 Mole ratio: 1 mol CaC 2 = 1 mol C 2 H 2 (from balanced equation) Molar Mass (MM) of CaC 2 = 64.0 g/mol CaC 2 Molar Mass of C 2 H 2 = 26.0g/mol C 2 H 2 What are we asked for? grams of C 2 H 2 produced CaC 2 + 2H 2 O C 2 H 2 + Ca(OH) 2 mass A moles A moles B mass B

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Learning check How many grams of O 2 are produced when a sample of 29.2 g of H 2 O is decomposed by electrolysis according to this balanced equation: 2H 2 O 2H 2 + O 2 CW p 393 #67-70

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Limiting Reactant and Percent Yield Limiting reactant (reagent): reactant that determines the amount of product that can be formed in the reaction. Excess reactant (reagent): reactant that is not completely used up in a reaction.

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Ex. 1 Copper reacts with sulfur to form copper (I) sulfide according to the following balanced equation. 2Cu + S Cu 2 S What is the limiting reactant when 80.0 g Cu react with 25.0g S? Calculate using stoichiometry how much product ( Cu 2 S) each amount of reactant produces respectively.

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Ex.2 The reaction between solid white phosphorus (P 4 ) and oxygen produces solid tetraphosphorus decoxide (P 4 O 10 ). Determine the mass of P 4 O 10 produced if 25.0g P 4 and 50.0g of oxygen are combined. How much excess reactant remains after the reaction stops?

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Theoretical yield: The maximum quantity of product that a reaction could theoretically make (calculated based upon limiting reactant through stoichiometry). Actual yield : The amount of product that was obtained experimentally. This is the amount you really got. Percent Yield = Actual Yield x 100 Theoretical Yield Percent Yield

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Ex. 1 ) 0.500 g of silver nitrate, AgNO 3, yields 0.455g of silver chromate, Ag 2 CrO 4, according to the following BALANCED equation. Calculate the percent yield of the reaction. 2AgNO 3 (aq)+ K 2 CrO 4 (aq) Ag 2 CrO 4 (s) + 2KNO 3 (aq)

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Stoichiometry Chapter 12.

Stoichiometry Chapter 12.

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