UNIT 9 STOICHIOMETRY

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 F +S+2W+H+2P  FSW2HP2 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 FSW2HP2. The balanced equation for the production of 1 bike is. F +S+2W+H+2P  FSW2HP2

F S 2W + H P FSW 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 FSW2HP2=2W (balanced eqn) What is unknown? # of wheels = ? wheels F +S+2W+H+2P  FSW2HP2

We can make the same kinds of connections from a chemical rxn eqn. The connection between wheels and bikes is 2 wheels per bike. We can use this information as a conversion factor to do the calculation. 2 W 640 FSW2HP2 = 1280 wheels 1 FSW2HP2 We can make the same kinds of connections from a chemical rxn eqn. N2(g) + 3H2(g)  2NH3(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 N2 reacts with 3 moles of H2 to produce 2 moles of NH3. N2 and H2 will always react to form ammonia in this 1:3:2 ratio of moles. So if you started with 10 moles of N2 it would take 30 moles of H2 and would produce 20 moles of NH3 N2(g) + 3H2(g)  2NH3(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.

The following rxn shows the synthesis of aluminum oxide. MOLE – MOLE EXAMPLE The following rxn shows the synthesis of aluminum oxide. 3O2(g) + 4Al(s)  2Al2O3(s) 3O2(g) + 4Al(s)  2Al2O3(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 Al2O3

MOLE – MOLE EXAMPLE Solve for the unknown: 3O2(g) + 4Al(s)  2Al2O3(s) 2 mol Al2O3 1.8 mol Al = 0.90mol Al2O3 4 mol Al Mole Ratio

MOLE – MOLE EXAMPLE 2 The following rxn shows the synthesis of aluminum oxide. 3O2(g) + 4Al(s)  2Al2O3(s) If you wanted to produce 24 mols of product how many mols of each reactant would you need? Given: 24 moles of Al2O3 Uknown: ____ moles of Al ____ moles of O2

MOLE – MOLE EXAMPLE 2 Solve for the unknowns: 3O2(g) + 4Al(s)  2Al2O3(s) 4 mol Al 24 mol Al2O3 = 48 mol Al 2 mol Al2O3 3 mol O2 24 mol Al2O3 = 36 mol O2 2 mol Al2O3

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.

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

MASS – MASS CALCULATIONS Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2). CaC2 + 2H2O  C2H2 + Ca(OH)2 How many grams of C2H2 are produced by adding water to 5.00 g CaC2?

MASS – MASS CALCULATIONS CaC2 + 2H2O  C2H2 + Ca(OH)2 What do we know? Given mass = 5.0 g CaC2 Mole ratio: 1 mol CaC2 = 1 mol C2H2 (from balanced equation) Molar Mass (MM) of CaC2 = 64.0 g/mol CaC2 Molar Mass of C2H2 = 26.0g/mol C2H2 What are we asked for? grams of C2H2 produced mass A  moles A  moles B  mass B

Learning check How many grams of O2 are produced when a sample of 29.2 g of H2O is decomposed by electrolysis according to this balanced equation: 2H2O  2H2 + O2 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  Cu2S What is the limiting reactant when 80.0 g Cu react with 25.0g S? Calculate using stoichiometry how much product (Cu2S) each amount of reactant produces respectively.

Ex.2 The reaction between solid white phosphorus (P4) and oxygen produces solid tetraphosphorus decoxide (P4O10). Determine the mass of P4O10 produced if 25.0g P4 and 50.0g of oxygen are combined. How much excess reactant remains after the reaction stops? CW p383 #23,24

Percent Yield 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

2AgNO3 (aq)+ K2CrO4 (aq)  Ag2CrO4 (s) + 2KNO3 (aq) Ex. ) 0.500 g of silver nitrate, AgNO3, yields 0.455g of silver chromate, Ag2CrO4, according to the following BALANCED equation. Calculate the percent yield of the reaction. 2AgNO3 (aq)+ K2CrO4 (aq)  Ag2CrO4 (s) + 2KNO3 (aq) CW p395 #90-92