Presentation on theme: "Unit 8 Stoichiometric Relationships In Chemical Reactions Mole Ratios Stoichiometry applied to gases Fractional Components of Chemical Samples Chemical."— Presentation transcript:
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.
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 a) total number of atoms during a chemical reaction does not change because matter cannot be created nor destroyed. b) atoms are rearranged.
Mole Ratios For the equation 2 H 2 + 1 O 2 > 2 H 2 O 2 moles of H 2 react with 1 mole of O 2 to get 2 moles of H 2 O. 4 + hydrogen atoms 2 yields oxygen atoms 2 water molecules [6 atoms] Furthermore, the Law of Conservation of Matter shows that
Q The salt potassium chlorate undergoes decomposition when heat is applied to form the salt potassium chloride and the gas oxygen. a)Write the balanced equation. b)What is the ratio of moles: potassium chlorate to potassium chloride to oxygen? 2 23 KClO 3 KCl + O 2 2 23
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 Information Given by the Chemical Equation
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.
2 CO + O2 > 2 CO2 2 moles CO + 1mole O2 = 2 moles CO2 Since 1 mole of CO = 28.01 g, 1 mole O2 = 32.00 g, and 1 mole CO2 = 44.01 g 2(28.01) g CO + 1(32.00) g O2 = 2(44.01) g CO2 Information Given by the Chemical Equation
Write the balanced equation 2 CO + O 2 > 2 CO 2 Use the coefficients to find the mole relationship 2 moles CO = 1 mol O 2 = 2 moles CO 2 Determine the Number of Moles of Carbon Monoxide required to react with 3.2 moles Oxygen, and determine the moles of Carbon Dioxide produced Example #1
®Use dimensional analysis 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
Write the balanced equation 2 CO + O 2 > 2 CO 2 Use the coefficients to find the mole relationship 2 moles CO : 1 mol O 2 : 2 moles CO 2 Determine the Molar Mass of each 1 mol CO = 28.01 g 1 mol O 2 = 32.00 g 1 mol CO 2 = 44.01 g 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 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 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
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? = mol O 2 1.17 = g KClO 3 3.35 KClO 3 KCl + O 2 2 23
Limiting And Excess Reagent Problems
8 meals cwx.prenhall.com Determine the number of packaged meals based on the given supply: 3 Excess sandwiches 2 Excess oranges One packaged meal for the USM service day:
In the chemical reaction Mg (s) + 2 HCl (aq) > MgCl 2 (aq) + H 2 (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.
Product For another chemical reaction C + O 2 CO 2 Initial supply of reactant
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 wont be told which one. 3.The following problem shows one way to determine the reactant that is in excess
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) This means that S is in excess. In other words, after the reaction there will be CaS, unreacted S, and no Ca. Higher.
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. Start with the limiting reagent since all of it will be used up. Look what happens if you start with the excess reactant: Ca(s) + S(s) CaS(s)
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
More Practice 4.00 g of calcium is mixed with 4.00 g of phosphorus and ignited to get solid calcium phosphide, Ca 3 P 2. a)Determine which reactant is in excess. (ans. P) b)How many grams of product were formed? (ans. 6.05 g) c)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)