STOICHIOMETRY

STOICHIOMETRY : 

STOICHIOMETRY USING THE REACTION EQUATION LIKE A RECIPE

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

USING EQUATIONS : 

Equations are a chemist’s recipe. Eqs 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 rxn, 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. USING EQUATIONS

USING EQUATIONS : 

The calculation of quantities in chem-ical reactions is called stoichiometry. Imagine you are in charge of manu-facturing for Rugged Rider Bicycle Company. The business plan for Rugged Rider requires the production of 128 custom-made bikes each day. You are responsible for insuring that there are enough parts at the start of each day. USING EQUATIONS

USING EQUATIONS : 

Assume that the major components of the 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. USING EQUATIONS F +S+2W+H+2P → FSW2HP2

USING EQUATIONS : 

Now in a 5 day workweek, Rugged Riders 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 USING EQUATIONS

Slide 8: 

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 FSW2HP2 1 FSW2HP2 2 W The key is the “coefficient ratio”.

Slide 9: 

Say I want to make a bacon double cheeseburger. Let’s get our recipe together. 1 hamburger bun + 2 hamburger patties + 2 slices of cheese + 4 strips of bacon = 1 bacon doublecheeseburger

Slide 10: 

How many bacon double cheeseburgers can you make if you start with: a. 1 bun, 2 patties, 2 slices of cheese, 4 strips of bacon b. 2 buns, 4 patties, 4 slices of cheese, 8 strips of bacon c. 1 dozen buns, 2 dozen patties, 2 dozen slices of cheese, 4 dozen strips of bacon d. 1 mole of buns, 2 moles of patties, 2 moles of cheese slices, 4 moles of bacon strips e. 10 buns, 20 patties, 2 slices of cheese, 40 strips of bacon

Slide 11: 

11 We can make the same kinds of connections from a chemical rxn eqn. N2(g) + 3H2(g) → 2NH3(g)

Slide 12: 

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

Slide 13: 

Using the coefficients, from the balan-ced rxn 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.

Slide 14: 

Using the coefficients of balanced rxn equations and our knowledge of mole conversions we can perform powerful calculations. A.K.A. stoichiometry. A balanced rxn equation is essential for all calculations involving amounts of reactants and products. If you know the number of moles of 1 substance, the balanced eqn allows you to calc. the number of moles of all other substances in a rxn equation.

MOLE – MOLE EXAMPLE : 

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

MOLE – MOLE EXAMPLE : 

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

MOLE – MOLE EXAMPLE 2 : 

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: 24 mol Al2O3 2 mol Al2O3 4 mol Al = 48 mol Al MOLE – MOLE EXAMPLE 2 3O2(g) + 4Al(s) → 2Al2O3(s) 24 mol Al2O3 2 mol Al2O3 3 mol O2 = 36 mol O2

MASS – MASS CALCULAT’NS : 

MASS – MASS CALCULAT’NS No lab balance measures moles directly, generally mass is the unit of choice. From the mass of 1 reactant or prod-uct, the mass of any other reactant or product in a given chemical equation can be calculated, provided you have a balanced rxn equation. As in mole-mole calcs, the unknown can be either a reactant or a product.

MASS – MASS CALCULAT’NS 1 : 

Acetylene gas (C2H2) is produced by adding water to calcium carbide (CaC2). How many grams of C2H2 are produced by adding water to 5.00 g CaC2? CaC2 + 2H2O → C2H2 + Ca(OH)2 MASS – MASS CALCULAT’NS 1

MASS – MASS CALCULAT’NS 1 : 

What do we know? Given mass = 5.0 g CaC2 Mole ratio: 1 mol CaC2 = 1 mol C2H2 MM of CaC2 = 64.0 g CaC2 MM of C2H2 = 26.0g C2H2 What are we asked for? grams of C2H2 produced MASS – MASS CALCULAT’NS 1

MASS – MASS CALCULAT’NS 1 : 

mass A  moles A  moles B  mass B 5.0 g CaC2 64.0 g CaC2 1 mol CaC2 MASS – MASS CALCULAT’NS 1 1mol CaC2 1 mol C2H2 1mol C2H2 26.0 g C2H2 = 2.03 g C2H2

MASS – MASS CALCULAT’NS 2 : 

You’ve recently learned that Copper will replace silver ions out of solution. You’re eyes light up with this money making opportunity. However, you decide it might be best if you did some preliminary calculations to determine to the feasibility of this get rich scheme. Copper is not very hard to find, however the largest size of Silver nitrate found in the Flinn Catalog is the 500 g size and it costs $305.91. Currently Silver sells for $9.00/ounce on the stock market. How much money could you sell your manufactured Silver for? MASS – MASS CALCULAT’NS 2

MASS – MASS CALCULAT’NS 2 : 

What do we know? Given mass = 500 g of AgNO3 Mole ratio: 2 mol AgNO3 = 2 mol Ag MM of AgNO3: 169.84g = 1mol MM of Ag: 107.87 g = 1mol Price of Silver: $9.00 = 1 ounce Conversion g to oz: 28.23g = 1 oz MASS – MASS CALCULAT’NS 2 Cu + 2AgNO3  2Ag + Cu(NO3)2 Cu + 2AgNO3  2Ag + Cu(NO3)2

MASS – MASS CALCULAT’NS 2 : 

500 g AgNO3 169.8gAgNO3 1mol AgNO3 MASS – MASS CALCULAT’NS 2 2 mol AgNO3 2 mol Ag 1mol Ag 107.87g Ag = $101.24 28.23 g 1 oz 1 oz $9.00

Slide 26: 

A balanced reaction equation indicates the relative numbers of moles of reactants and products. We can expand our stoichiometric calculations to include any unit of measure that is related to the mole. The given quantity can be expressed in numbers of particles, units of mass, or volumes of gases at STP. The problems can include mass-volume, volume-volume, and particle-mass calculations.

Slide 27: 

In any of these problems, the given quantity is first converted to moles. Then the mole ratio from the balanced eqn is used to convert from the moles of given to the number of moles of the unknown Then the moles of the unknown are converted to the units that the problem requests. The next slide summarizes these steps for all typical stoichiometric problems

MORE MOLE EXAMPLES : 

MORE MOLE EXAMPLES How many molecules 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

MORE MOLE EXAMPLES : 

What do we know? Mass of H2O = 29.2 g H2O 2 mol H2O = 1 mol O2 (from balanced equation) MM of H2O = 18.0 g H2O 1 mol O2 = 6.02x1023 molecules of O2 What are we asked for? molecules of O2 MORE MOLE EXAMPLES

Slide 31: 

mass A  mols A  mols B  molecules B 29.2 g H2O 18.0 g H2O 1 mol H2O 2 mol H2O 1 mol O2 1 mol O2 6.02x1023 molecules O2 = 4.88 x 1023 molecules O2

MORE MOLE EXAMPLES : 

The last step in the production of nitric acid is the reaction of NO2 with H2O. 3NO2+H2O→2HNO3+NO How many liters of NO2 must react with water to produce 5.00x1022 molecules of NO? MORE MOLE EXAMPLES

MORE MOLE EXAMPLES : 

What do we know? Molecules NO = 5.0x1022 molecules NO 1 mol NO = 3 mol NO2 (from balanced equation) 1 mol NO = 6.02x1023 molecules NO 1 mol NO2 = 22.4 L NO2 What are we asked for? Liters of NO2 MORE MOLE EXAMPLES

Slide 34: 

molecules A mols mols B volume B 5.0x1022 mol-ecules NO 6.02x1023 mol-ecules NO 1 mol NO 1 mol NO 3 mol NO2 = 5.58 L NO2 1 mol NO2 22.4 L NO2

Slide 35: 

Aspirin can be made from a chemical rxn between the reactants salicylic acid and acetic anhydride. The products of the rxn are acetyl-salicylic acid (aspirin) and acetic acid (vinegar). Our factory makes 125,000 100-count bottles of Bayer Aspirin/day. Each bottle contains 100 tablets, and each tablet contains 325mg of aspirin. How much in kgs + 10% for production problems, of each reactant must we have in order to meet production?

Slide 36: 

What do we know? Make 125,000 aspirin bottles/day 100 aspirin/bottle 325 mg aspirin/tablet Mole ratio of aspirin to salicylic acid (1:1) and acetic anhydride (1:1) MM aspirin = 180.11g MM C7H6O3 = 138.10g MM C4H6O3 = 102.06g What are we asked for? Mass of salicylic acid in kgs + 10% Mass of acetic anhydride in kgs + 10%

Slide 37: 

125,000 bottles 1 bottle 100 tablets 1 tablet 325mg asp. 1000 mg 1 g = 22,549.4 mols aspirin 180.16g 1mol asp.

Salicylic Acid: : 

22,549.4 mols aspirin 1 mol asp 1 mol C7H6O3 1 mol C7H6O3 136.10g C7H6O3 1000 g 1 kg = 3068.97 kg salicylic acid + (306.897 g) Salicylic Acid: = 3380 kg of salicylic acid

Acetic Anhydride: : 

22,549.4 mols aspirin 1 mol asp 1 mol C4H6O3 1 mol C4H6O3 102.06g C4H6O3 1000 g 1 kg = 2301.39 kg Acetic anhydride + 230.139 kg Acetic Anhydride: = 2530 kg Acetic anhydride

How many grams of silver will be formed from 12.0 g copper? : 

How many grams of silver will be formed from 12.0 g copper? Cu + 2AgNO3 → 2Ag + Cu(NO3)2

Slide 41: 

1. The human body needs at least 1.03 x 102 mol O2 every minute. If all of this oxygen is used for the cellular respiration reaction that breaks down glucose, how many grams of glucose does the human body consume each minute? C6H12O6(s) + 6 O2(g) -----> 6 CO2(g) + 6 H2O(l)

Slide 42: 

2. In the space shuttle, the CO2 that the crew exhales is removed from the air by a reaction within canisters of lithium hydroxide. On average, each astronaut exhales about 20.0 mol of CO2 daily. What volume of water will be produced when this amount of CO2 reacts with an excess of LiOH? CO2(g) + 2 LiOH(s) ----> Li2CO3(aq) + H2O

Slide 43: 

3. Carbon monoxide can be combined with hydrogen to produce methanol, CH3OH. Methanol is used as an industrial solvent, as a reactant in synthesis, and as a clean-burning fuel for some racing cars. If you had 152.5 kg CO and 24.50 kg H2, how many kilograms of CH3OH could be produced?