# USING EQUATIONS  Nearly everything we use is manufactured from chemicals. › Soaps, shampoos, conditioners, cd’s, cosmetics, medications, and clothes.

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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.

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.

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

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

 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”.

◦ 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)

 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.

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)

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

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

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

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

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.

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)

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

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

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

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.

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.

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?

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

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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