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Section 6.5—Stoichiometry

Section 6.5—Stoichiometry. Can we determine, in the lab, the concentration of electrolytes?. Objectives Perform mole to mole stoichiometric calculations using mole ratios from balanced chemical equations

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Section 6.5—Stoichiometry

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  1. Section 6.5—Stoichiometry Can we determine, in the lab, the concentration of electrolytes? • Objectives • Perform mole to mole stoichiometric calculations using mole ratios from balanced chemical equations • Explain and use titrations, indicators, and stoichiometry to determine the concentration of an unknown acid

  2. What is stoichiometry? • A chemist knows that a chemical reaction stops when one of the reactants is used up. • Therefore, the amounts of other reactants used or products formed depends on the amount of reactant that is used up. • Stoichiometric calculations allow us to use this information to determine the amounts of other reactants used or products formed in a reaction.

  3. Stoichiometry is the study of quantitative relationships between amounts of reactants used and the products formed by a chemical reaction. • It is based on the Law of Conservation of Mass: the amount of matter present at the end of a reaction is the same as was present at the beginning. • The total mass of the reactants equals the mass of the products.

  4. Interpretation of Chemical Equations • The coefficients in a chemical equation • can be interpreted several ways. • We have learned to interpret them in • terms of representative particles.

  5. Coefficients can also represent “number of moles”! We can, therefore, interpret coefficients as numbers of moles!

  6. What do those coefficients really mean? For every 2 moles of H2… and 2 moles of H2O are produced 2 2 2 H2 + O2 2 H2O No coefficient = 1 1 mole of O2 is need to react…

  7. Mole Ratios • In the following reaction, how many moles of Al will react with 3 moles of Br2? How many moles of AlBr3 are formed? • 2Al(s) + 3Br2(l) → 2AlBr3(s) • The relationships above can be expressed as MOLE RATIOS. • A mole ratio is a ratio between the numbers of moles of any two substances in a balanced chemical equation.

  8. Mole Ratios • 2Al(s) + 3Br2(l) → 2AlBr3(s) • For example, the mole ratios between Al and AlBr3 are: 2 mol Al and 2 mol AlBr3 2 mol AlBr3 2 mol Al

  9. Mole Ratios • Mole ratios are used as conversion factors in stoichiometric calculations • With the chemical equation and the mole ratios, you can calculate the amount used of any reactant in the equation and the maximum amount of product you can obtain.

  10. Mole to Mole Conversions Problem: Potassium reacts with water to produce potassium hydroxide and hydrogen gas. Determine the number of moles of hydrogen gas produced when 0.04 mol of K is used? 2K(s) + 2H2O(l)  2KOH(aq) + H2(g)

  11. Mole to Mole Conversions • Step 1: Write a balanced equation if you are not given one. 2K(s) + 2H2O(l) 2KOH(aq) + H2(g) • Step 2: Identify the known/given and the unknown in the problem. • 0.04 mol K is given; mol H2 is unknown. • Step 3: Use the balanced equation to determine the mole ratio in which mol unknown is the numerator and mol known is the denominator. mol unknown = 1 mol H2 mol known 2 mol K

  12. Mole to Mole Conversions • Step 4: Multiply the known by the mole ratio. Note that known substance & its units will cancel if it is correctly done. 0.04 mol K x 1 mol H2 = 0.02 mol H2 2 mol K We have gone from moles of known to moles of unknown!

  13. Practice Problems • How many moles of carbon dioxide gas are produced when 10.0 mol of propane (C3H8) are burned in an excess of oxygen? C3H8 + 5 O2 3CO2 + 4H2O • If 4.2 mole of H2 reacts completely with O2, how many moles of O2 are needed? 2 H2 + O2 2 H2O

  14. But we can’t measure moles in lab! We can’t go to the lab and count or measure moles…so we need a way to convert from moles to units we CAN measure in the lab, such as grams and liters! Remember molar mass?? It is the number of grams in 1 mole of a substance. Using molar mass, we can convert from moles to a mass!

  15. Stoichiometry with Moles & Mass Example: How many grams of AgCl will form if 0.45 mole AgNO3 is reacted as follows: 2 AgNO3 + CaCl2 2 AgCl + Ca(NO3)2

  16. Stoichiometry with Moles & Mass Example: How many grams of AgCl will form if 0.45 mole AgNO3 is reacted as follows: 2 AgNO3 + CaCl2 2 AgCl + Ca(NO3)2 Step 1: Use stoichiometry to find mol AgCl from 0.45 mol AgNO3 0.45 mol AgNO3 x 2 mol AgCl = 0.45 mol AgCl 2 mol AgNO3

  17. Stoichiometry with Moles & Mass Step 2: Convert the mol AgCl to grams using the molar mass 1 Ag = 107.9 0.45 mol AgCl x 143.4 g = 65 g 1 Cl = 35.5 1 mol 143.4

  18. Stoichiometry with Moles & Mass Example: How many grams of Ba(OH)2 are formed from 5.2 mol of NaOH in the following reaction: 2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl

  19. Stoichiometry with Moles & Mass Example: How many grams of Ba(OH)2 are formed from 5.2 mol of NaOH in the following reaction: 2 NaOH + BaCl2 Ba(OH)2 + 2 NaCl Step 1: Go from 5.2 mol NaOH to moles Ba(OH)2 5.2 mol NaOH X 1 mol Ba(OH)2 = 2.6 mol Ba(OH)2 2 mol NaOH

  20. Stoichiometry with Moles & Mass Step 2: Convert 2.6 mol Ba(OH)2 to mass Ba = 137.3 2.6 mol Ba(OH)2 X 171.3 g = 445 g 2 O = 32.0 1 mol 2 H = 2.0 171.3

  21. Practice Problems • 4Fe + 3O2 2Fe2O3 • How many grams of iron must be used to • produce 3.0 mol of Fe2O3? • 2. How many grams of oxygen are required to react with 4.2 mol of Fe? • How many grams of Fe2O3 are formed when 0.85 mol of Fe are oxidized? • 4. How many grams of oxygen are used to produce 1.8 mol of Fe2O3?

  22. What about gases? One mole of any gas (at the standard temperature of 0 oC and pressure of 1 atm) will occupy a volume of 22.4 L. So, for gases, we can determine the volume of a gas from its number of moles. This is sometimes more helpful to scientists than the mass of the gas!!

  23. Stoichiometry with Gases Example: If 1.5 mol of zinc reacted, what volume of H2 gas will be produced at STP? 2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g)

  24. Stoichiometry with Gases Example: If 1.5 mol of zinc reacted, what volume of H2 gas will be produced at STP? 2 HCl (aq) + Zn (s)  ZnCl2 (aq) + H2 (g) Step 1: Go from 1.5 mol Zn to mol H2 1.5 mol Zn x 1 mol H2 = 1.5 mol H2 1 mol Zn

  25. Stoichiometry with Gases Step 2: Convert 1.5 mol H2 into Liters 1.5 mol H2 x 22.4 L = 33.6 L H2 1 mol

  26. Practice • 2H2 (g) + O2 (g)  2H2O (l) • How many liters of hydrogen are required to produce 8.4 mol of water? • How many liters of oxygen are required to produce 0.25 mol of water?

  27. Keeping all these equalities straight!

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