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Unit 8 Stoichiometric Relationships In Chemical Reactions Mole Ratios

Unit 8 Stoichiometric Relationships In Chemical Reactions Mole Ratios Stoichiometry applied to gases Fractional Components of Chemical Samples Chemical Reactions Give Off or Take In Energy Limiting And Excess Reagent Problems.

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Unit 8 Stoichiometric Relationships In Chemical Reactions Mole Ratios

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  1. Unit 8 Stoichiometric Relationships In Chemical Reactions Mole Ratios Stoichiometry applied to gases Fractional Components of Chemical Samples Chemical Reactions Give Off or Take In Energy Limiting And Excess Reagent Problems Works best when seen as a slide show. Click the ‘Slide Show’ button on the lower right.

  2. Mole Ratios • The coefficients of a chemical equation represent a ratio of moles. They are used to balance the equation thereby reflecting the Law of Conservation of Matter. • This law states that the • total number of atoms during a chemical reaction does not change because matter cannot be created nor destroyed. • atoms are rearranged.

  3. Mole Ratios For the equation 2 H2 + 1 O2 ———> 2 H2O 2 moles of H2 react with 1 mole of O2 to get 2 moles of H2O. Furthermore, the Law of Conservation of Matter shows that 2 water molecules [6 atoms] 4 + hydrogen atoms 2 yields oxygen atoms

  4. 2 2 3 3 2 2 • Q The salt potassium chlorate undergoes decomposition when heat is applied to form the salt potassium chloride and the gas oxygen. • Write the balanced equation. • What is the ratio of moles: potassium chlorate to potassium chloride to oxygen? KClO3 KCl + O2

  5. Information Given by theChemical Equation • The coefficients in the balanced chemical equation shows the molecules and mole ratio of the reactants and products • Since moles can be converted to masses, we can determine the mass ratio of the reactants and products as well

  6. Stoichiometry Rules! • Write a chemical equation. • Balance the equation! • Determine start and destination for dimensional analysis. • Convert start details into moles. • Multiply by the appropriate mole ratio. • Convert to correct units for destination.

  7. Example #1 • Write the balanced equation 2 CO + O2 —> 2 CO2 • Use the coefficients to find the mole relationship 2 moles CO = 1 mol O2 = 2 moles CO2 Determine the Number of Moles of Carbon Monoxide required to react with 3.2 moles Oxygen, and determine the moles of Carbon Dioxide produced

  8. Example #1 Determine the Number of Moles of Carbon Monoxide required to react with 3.2 moles Oxygen, and determine the moles of Carbon Dioxide produced • Use dimensional analysis

  9. Example #2 Determine the Number of grams of Carbon Monoxide required to react with 48.0 g Oxygen, and determine the mass of Carbon Dioxide produced • Use the molar mass of the given quantity to convert it to moles • Use the mole relationship to convert the moles of the given quantity to the moles of the desired quantity

  10. Example #2 Determine the Number of grams of Carbon Monoxide required to react with 48.0 g Oxygen, and determine the mass of Carbon Dioxide produced Use the molar mass of the desired quantity to convert the moles to mass

  11. KClO3 KCl + O2 2 3 2 • If 0.783 mole of potassium chlorate decomposed, how many moles of oxygen form? • If 2.04 grams of potassium chloride were produced, how many grams of potassium chlorate must have decomposed? 1.17 = mol O2 = g KClO3 3.35

  12. Limiting And Excess Reagent Problems

  13. 8 meals One packaged meal for the USM service day: Determine the number of packaged meals based on the given supply: 2 Excess oranges 3 Excess sandwiches cwx.prenhall.com

  14. In the chemical reaction Mg(s) + 2 HCl(aq) ———> MgCl2(aq) + H2(g) performed in many high school laboratories, more HCl is usually used than is needed. This is done to assure that all the magnesium reacts. The HCl, therefore, is in excess and some of it is left over when the reaction stops. No magnesium will be present once the reaction is over.

  15. For another chemical reaction C + O2 CO2 Product Initial supply of reactant

  16. 1. In many chemistry problems you are told which reactant is in excess and solve based on the specified quantities of the other reactant. 2. In some instances, however, you will be given quantities of both reactants. One is probably in excess, but you won’t be told which one. 3. The following problem shows one way to determine the reactant that is in excess

  17. This means that S is in excess. In other words, after the reaction there will be CaS, unreacted S, and no Ca. Q 5.00 g of calcium reacts with 5.00 g of sulfur in a composition reaction. Do a calculation to find the a) Excess reactant. Ca(s) + S(s) CaS(s) Higher.

  18. Q 5.00 g of calcium reacts with 5.00 g of sulfur in a composition reaction. Do a calculation to find the b) Grams of product formed. Ca(s) + S(s) CaS(s) Start with the limiting reagent since all of it will be used up. Look what happens if you start with the excess reactant:

  19. c) Calculate the grams of excess reactant unreacted once the reaction is completed. 10.0 g of original reaction mixture – 9.025 g of CaS formed = 1.0 g S left over

  20. More Practice • 4.00 g of calcium is mixed with 4.00 g of phosphorus and ignited to get solid calcium phosphide, Ca3P2. • Determine which reactant is in excess. (ans. P) • How many grams of product were formed? (ans. 6.05 g) • How many grams of the excess reactant were left over once the reaction was completed? • (8 grams reactants – 6.05 grams calcium phosphide = 1.95 grams phosphorus left)

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