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Unit 8 - Stoichiometry. 2CuCl + H 2 S → Cu 2 S +2HCl The coefficients in the equation can either mean particles or moles.

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Presentation on theme: "Unit 8 - Stoichiometry. 2CuCl + H 2 S → Cu 2 S +2HCl The coefficients in the equation can either mean particles or moles."— Presentation transcript:

1 Unit 8 - Stoichiometry

2 2CuCl + H 2 S → Cu 2 S +2HCl The coefficients in the equation can either mean particles or moles.

3 Stoichiometry- is the study of the amount of substances involved in a chemical reaction. SO YOU MUST HAVE A BALANCED EQUATION TO DO THESE PROBLEMS!!

4 Molar Mass- The mass of one mole. Molar Volume- The volume of one mole of any gas at STP(standard temperature and pressure= 0 ⁰C (273 K) and 1 atm of pressure) and is 22.4 L.

5 Molarity – the number of moles per liter of solution. Mole ratio – is the ratio of moles of one element to another in a chemical reaction. The mole ratio can be used in a problem to find the amount of one substance in an equation when another is known.

6 A mole ratio is a conversion factor that relates the amounts in moles of 2 substances involved in a chemical reaction. The mole ratio comes directly from the balanced chemical equation. 2CuCl + H 2 S → Cu 2 S + 2HCl 2 mol CuCl = 1 mol Cu 2 S in this equation

7 The coefficients in the equation represent the relative amounts of moles of reactants and products. 2Al 2 O 3 → 4Al + 3O 2 6 mole ratios can be written for this chemical equation. See if you can name all 6 ratios.

8 Mole Energy Ratios – The ratio of moles to energy in a chemical reaction. Energy is written in some chemical equations. However, energy is required for all chemical reactions.

9 The heat of reaction (H rxn ) is the quantity of energy released or absorbed as heat during a chemical reaction. You can think of heat of reaction as the difference between the stored energy of the reactants and the products.

10 Activation energy, E a, is the energy required for a reaction to occur.

11 2H 2 + O 2 → 2H 2 O + 483.6 kJ This equation is an example of a thermochemical equation, an equation that includes a quantity of energy released or absorbed as heat during a chemical reaction.

12 2H 2 + O 2 → 2H 2 O + 483.6 kJ The quantity of heat released during the formation of water from H 2 and O 2 Is proportional to the quantity of water formed. Producing 2Xs as much water vapor would require twice as many moles of reactants and would release 2Xs the amount of energy, 967.2 kJ. H rxn = H products - H reactants

13 Endothermic – means the reaction absorbs more energy from the surroundings than it releases so it feel cold. H rxn is positive so the H products greater than the H reactants. Endothermic reactions are less common than exothermic ones.

14 H rxn of an Endothermic Reactions

15 Exothermic – means the reactions releases more energy to the surrounding than it absorbs so it feels hot. Combustion reactions are exothermic reactions. In exothermic reaction, H rxn is negative since reactants have more energy than the products.

16 H rxn of an Exothermic Reaction

17 The same methods used to solve stoichiometry problems can be used with thermochemical equations. Essentially what we do is to treat the quantity of heat as if it were a quantity of a reactant or product. Consider the following example. What is the enthalpy change associated with the formation of 95.6 g of HCl(g) in the reaction? H 2 (g) + Cl 2 (g) → HCl(g) ΔH = -184.6 kJ

18 Solution The calculation involves converting grams of HCl to moles of HCl, and then to kilojoules of heat. H 2 (g) + Cl 2 (g) → 2HCl(g) H rxn = -184.6 kJ 95.6 g HCl ÷ 36.46 g/1mol HCl = 2.62 mol HCl 95.6 g. X = - 240 KJ

19 Thermochemistry Problem 2 Compute the heat of reaction for the formation of 1.99 g. of Na 2 O 2. 2 Na 2 O 2 + 2H 2 O → 4NaOH + 215.76 kJ 1.99 g

20 Percent Yield Percentage yield is the difference between how much product is actually made in the lab versus how much product theoretically can be made. % Yield = actual yield ÷ theoretical yield X 100

21 Limiting Reactants The Limiting Reactant is the reactant that is completely consumed in the reaction. All other reactants are in excess. (some left over)

22 To calculate from masses given for more than one reactant: 1.Calculate the product using both masses given. 2. The reactant that produces the least amount of product is the limiting reactant and that is the amount of product that can be produced!


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