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Stoichiometry

Stoichiometry. Calculating quantities using balanced chemical equations. +. +. +. +. Using Equations. Everyday Equations. Travel Time Tricycle company produces 640 tricycles each week. How can you determine the number of parts they need per week?. +. +. +. +. Using Equations.

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Stoichiometry

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  1. Stoichiometry Calculating quantities using balanced chemical equations

  2. + + + + Using Equations Everyday Equations • Travel Time Tricycle company produces 640 tricycles each week. How can you determine the number of parts they need per week?

  3. + + + + Using Equations Everyday Equations • Travel Time Tricycle company produces 640 tricycles each week. How can you determine the number of parts they need per week? • The major components are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P).

  4. + + + + Using Equations Everyday Equations • Travel Time Tricycle company produces 640 tricycles each week. How can you determine the number of parts they need per week? • The balanced equation for making a single tricycle is: • F + S + 3W + H + 2P  FSW3HP2

  5. Using a Balanced Equation as a Recipe • In a five-day workweek, Travel Time is scheduled to make 640 tricycles. How many wheels should be in the plant on Monday morning to make these tricycles?

  6. 1 KNOWNS number of tricycles = 640 tricycles = 640 FSW3HP2 F + S + 3W + H + 2P  FSW3HP2 UNKNOWN number of wheels = ? wheels Analyze List the knowns and the unknown. Use the balanced equation to identify a conversion factor that will allow you to calculate the unknown. The conversion you need to make is from tricycles (FSW3HP2) to wheels (W).

  7. 2 3 W 3 W 1 FSW3HP2 and 1 FSW3HP2 1 FSW3HP2 3 W 640 FSW3HP2 = 1920 W When using conversion factors, remember to cross out like units when they are in both the numerator and denominator. This will help you check that you are using the correct conversion factor. Calculate Solve for the unknown. The desired unit is W; so use the conversion factor on the left. Multiply the number of tricycles by the conversion factor.

  8. Using Equations Balanced Chemical Equations Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or how much product will be formed in a reaction. • 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 reaction. This is called stoichiometry.

  9. Chemical Equations • Ammonia is produced industrially by the reaction of nitrogen with hydrogen. N2(g) + 3H2(g)  2NH3(g) • The balanced chemical equation tells you the relative amounts of reactants and product in the reaction. • Your interpretation of the equation depends on how you quantify the reactants and products.

  10. 2 atoms N + 6 atoms H 2 atoms N and 6 atoms H Chemical Equations • Number of Atoms At the atomic level, a balanced equation indicates the number and types of atoms that are rearranged to make the product or products. • In the synthesis of ammonia, the reactants are composed of two atoms of nitrogen and six atoms of hydrogen.

  11. Chemical Equations • Moles Since a balanced chemical equation tells you the number of representative particles, it also tells you the number of moles. • In the synthesis of ammonia, one mole of nitrogen molecules reacts with three moles of hydrogen molecules to form two moles of ammonia molecules. 1 mol N2 + 3 mol H2 2 mol NH3

  12. ( ) ( ) ( ) 6.02  1023 molecules N2 6.02  1023 molecules H2 6.02  1023 molecules NH2 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L Chemical Equations • The table below summarizes the information derived from the balanced chemical equation for the formation of ammonia.

  13. Writing and Using Mole Ratios A mole ratio is a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles. *Used to convert between a given number of moles of a reactant or product to moles of a different reactant or product

  14. 1 mol N2 2 mol NH3 3 mol H2 3 mol H2 1 mol N2 2 mol NH3 Writing and Using Mole Ratios • The three mole ratios derived from the balanced equation above are: • Look again at the balanced equation for the production of ammonia. • N2(g) + 3H2(g)  2NH3(g)

  15. Calculating Moles of a Product • How many moles of NH3 are produced when 0.60 mol of nitrogen reacts with hydrogen?

  16. 1 Analyze List the known and the unknown. N2 + 3H2 2NH3 • The conversion is mol N2 mol NH3. • 1 mol N2 combines with 3 mol H2 to produce 2 mol NH3. • To determine the number of moles of NH3, the given quantity of N2 is multiplied by the form of the mole ratio from the balanced equation that allows the given unit to cancel. KNOWN moles of nitrogen = 0.60 mol N2 UNKNOWN moles of ammonia = ? mol NH3

  17. 2 2 mol NH3 0.60 mol N2 = 1.2 mol NH3 1 mol N2 Remember that the mole ratio must have N2 on the bottom so that the mol N2 in the mole ratio will cancel with mol N2 in the known. Calculate Solve for the unknown. Multiply the given quantity of N2 by the mole ratio in order to find the moles of NH3.

  18. Writing and Using Mole Ratios Mass-Mass Calculations In the laboratory, the amount of a substance is usually determined by measuring its mass in grams. • If a given sample is measured in grams, then the mass can be converted to moles by using the molar mass. • Then the mole ratio from the balanced equation can be used to calculate the number of moles of the unknown. • If it is the mass of the unknown that needs to be determined, the number of moles of the unknown can be multiplied by the molar mass.

  19. Calculating the Mass of a Product • Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. The balanced equation is: • N2(g) + 3H2(g)  2NH3(g)

  20. 1 Analyze List the knowns and the unknown. N2(g) + 3H2(g)  2NH3(g) KNOWNS mass of hydrogen = 5.40 g H2 3 mol H2= 2 mol NH3 (from balanced equation) 1 mol H2 = 2.0 g H2 (molar mass) 1 mol NH3 = 17.0 g NH3 (molar mass) UNKNOWN mass of ammonia = ? g NH3

  21. 1 Analyze List the knowns and the unknown. • The mass of hydrogen will be used to find the mass of ammonia: g H2 g NH3. • The coefficients of the balanced equation show that 3 mol H2 reacts with 1 mol N2 to produce 2 mol NH3. • The following steps are necessary to determine the mass of the ammonia: g H2 mol H2 mol NH3 g NH3

  22. 2 1 mol H2 5.40 g H2 2.0 g H2 Given quantity Change given unit to moles Don’t forget to cancel the units at each step. Calculate Solve for the unknown. Start with the given quantity, and convert from mass to moles.

  23. 2 1 mol H2 2 mol NH3 5.40 g H2  2.0 g H2 3 mol H2 Given quantity Change given unit to moles Mole ratio Calculate Solve for the unknown. Then convert from moles of reactant to moles of product by using the correct mole ratio.

  24. 2 1 mol H2 2 mol NH3 17.0 g NH3 5.40 g H2   2.0 g H2 3 mol H2 1 mol NH3 Change moles to grams Given quantity Change given unit to moles Mole ratio Calculate Solve for the unknown. Finish by converting from moles to grams. Use the molar mass of NH3. = 31 g NH3

  25. 3 Evaluate Does the result make sense? • Because there are three conversion factors involved in this solution, it is more difficult to estimate an answer. • Because the molar mass of NH3 is substantially greater than the molar mass of H2, the answer should have a larger mass than the given mass. • The answer should have two significant figures.

  26. Phosphorus burns in air to produce a phosphorus oxide in the following reaction: • 4P(s) + 5O2(g)  P4O10(s) • What mass of phosphorus will be needed to produce 3.25 mol of P4O10?

  27. 3.25 mol P4O10  = 403 g P 31.0 g P 4 mol P 1 mol P 1 mol P4O10 • Phosphorus burns in air to produce a phosphorus oxide in the following reaction: • 4P(s) + 5O2(g)  P4O10(s) • What mass of phosphorus will be needed to produce 3.25 mol of P4O10?

  28. Other Stoichiometric Calculations Other Stoichiometric Calculations • What is the general procedure for solving a stoichiometric problem?

  29. Other Stoichiometric Calculations • 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. • Finally, the moles are converted to any other unit of measurement related to the unit mole, as the problem requires.

  30. molar mass and 1 mol 1 mol molar mass Other Stoichiometric Calculations The mole-mass relationship gives you two conversion factors.

  31. 1 mol 6.02x1023 particles and 6.02x1023 particles 1 mol 1 mol 22.4 L and 22.4 L 1 mol Other Stoichiometric Calculations • Recall that the mole can be related to other quantities: • 1 mol = 6.02 x 1023 particles • 1 mol of a gas = 22.4 L at STP • These provide four more conversion factors:

  32. Volume-Volume Stoichiometric Calculations • Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide, which contributes to photochemical smog. How many liters of nitrogen dioxide are produced when 34 L of oxygen react with an excess of nitrogen monoxide? Assume conditions are at STP. • 2NO(g) + O2(g)  2NO2(g)

  33. 1 Analyze List the knowns and the unknown. For gaseous reactants and products at STP, 1 mol of a gas is equal to 22.4 L. KNOWNS volume of oxygen = 34 L O2 2 mol NO2/1 mol O2 (mole ratio from balanced equation) 1 mol O2 = 22.4 L O2 (at STP) 1 mol NO2 = 22.4 L NO2 (at STP) UNKNOWN volume of nitrogen dioxide = ? L NO2

  34. 2 1 mol O2 34 L O2 22.4 L O2 Given quantity Change to moles Did you notice that the 22.4 L/mol factors canceled out? This will always be true in a volume-volume problem. Calculate Solve for the unknown. Start with the given quantity, and convert from volume to moles by using the volume ratio.

  35. 2 1 mol O2 2 mol NO2 22.4 L O2 34 L O2  1 mol O2 Given quantity Change to moles Mole ratio Calculate Solve for the unknown. Then, convert from moles of reactant to moles of product by using the correct mole ratio.

  36. 2 1 mol O2 22.4 L NO2 2 mol NO2 22.4 L O2 34 L O2   1 mol NO2 1 mol O2 Given quantity Change to moles Mole ratio Change to liters Calculate Solve for the unknown. Finish by converting from moles to liters. Use the volume ratio. = 68 L NO2

  37. 3 Evaluate Does the result make sense? • Because 2 mol NO2 are produced for each 1 mol O2 that reacts, the volume of NO2 should be twice the given volume of O2. • The answer should have two significant figures.

  38. Finding the Volume of a Gas Needed for a Reaction • Assuming STP, how many milliliters of oxygen are needed to produce 20.4 mL SO3 according to this balanced equation? • 2SO2(g) + O2(g)  2SO3(g)

  39. 1 Analyze List the knowns and the unknown. • For a reaction involving gaseous reactants or products, the coefficients also indicate relative amounts of each gas. • You can use the volume ratios in the same way you have used mole ratios. KNOWNS volume of sulfur trioxide = 20.4 mL 2 mL SO3/1 mL O2 (volume ratio from balanced equation) UNKNOWN volume of oxygen = ? mL O2

  40. 2 1 mL O2 20.4 mL SO3 = 10.2 mL O2 2 mL SO3 The volume ratio can be written using milliliters as the units instead of liters. Calculate Solve for the unknown. Multiply the given volume by the appropriate volume ratio.

  41. 3 Evaluate Does the result make sense? • Because the volume ratio is 2 volumes SO3 to 1 volume O2, the volume of O2 should be half the volume of SO3. • The answer should have three significant figures.

  42. Methane burns in air by the following reaction: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) What volume of water vapor is produced at STP by burning 501 g of methane?

  43. 1 mol CH4 2 mol H2O 22.4 L H2O 501 g CH4   16.05 g CH4 1 mol CH4 1 mol H2O • Methane burns in air by the following reaction: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) What volume of water vapor is produced at STP by burning 501 g of methane? = 1.40  103 L H2O

  44. Limiting and Excess Reagents • To make tacos, you need enough meat, cheese, lettuce, tomatoes, sour cream, salsa, and seasonings. • If you have only 2 taco shells, the quantity of taco shells will limit the number of tacos you can make. • Thus, the taco shells are the limiting ingredient.

  45. Limiting and Excess Reagents • In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.

  46. Limiting and Excess Reagents • A balanced chemical equation is a chemist’s recipe. • What would happen if two molecules (moles) of N2 reacted with three molecules (moles) of H2?

  47. Limiting and Excess Reagents • In this reaction, only the hydrogen is completely used up. • H2 is the limiting reagent, or the reactant that determines the amount of product that can be formed by a reaction.

  48. Limiting and Excess Reagents • The reactant that is not completely used up in a reaction is called the excess reagent. • In this example, nitrogen is the excess reagent because some nitrogen remains unreacted.

  49. Determining the Limiting Reagent in a Reaction • Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation: • 2Cu(s) + S(s)  Cu2S(s) • What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S?

  50. 1 Analyze List the knowns and the unknown. The number of moles of each reactant must first be found. The balanced equation is used to calculate the number of moles of one reactant needed to react with the given amount of the other reactant. KNOWNS UNKNOWN mass of copper = 80.0 g Cu mass of sulfur = 25.0 g S molar mass of Cu = 63.5 g/mol molar mass of S = 32.1 g/mol 1 mol S/2 mol Cu limiting reagent = ?

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