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2Mg (s) + O 2 → 2MgO INTERPRETING A CHEMICAL EQUATION Quantitative Interpretation of Chemical Reactions Stoichiometry is one of the most important topic.

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Presentation on theme: "2Mg (s) + O 2 → 2MgO INTERPRETING A CHEMICAL EQUATION Quantitative Interpretation of Chemical Reactions Stoichiometry is one of the most important topic."— Presentation transcript:

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2 2Mg (s) + O 2 → 2MgO

3 INTERPRETING A CHEMICAL EQUATION Quantitative Interpretation of Chemical Reactions Stoichiometry is one of the most important topic in chemistry. It involves the use of the chemical formulas, mole calculations, and chemical equations. Stoichiometry is also essential in industry, there, it is used to do cost and analysis for manufacturing chemicals.

4 STOICHIOMETRY : The relationship of quantities (mass of substance or volume of gas ) in a chemical change according to the balanced chemical equation.

5 LAW OF COMBINING VOLUMES The principle that volume of gases that combine in a chemical reaction, at the same temperature and pressure, are in the ratio of small whole numbers, also called Gay-Lussac’s Law of combining volumes.

6 H 2 + Cl 2 → 2HCl + →

7 That is, Hydrogen gas + chlorine → hydrogen chloride gas 1 volume 1 volume 2 volumes Ex. 50.0 ml 50.0 ml 100.0 ml Avogadro proposed that equal volumes of gas, at the same temperature and pressure, contain the same number of molecules. That is any two gases containing the same numbers of molecules, will occupy equal volumes.

8 That is: Hydrogen + oxygen → water 2 volumes 1 volume 2 volume 100.0 ml 50.0 ml 100.0 ml We can also look at the reaction of hydrogen and oxygen molecules 2 H 2 + O 2 → 2 H 2 O

9 Example Exercises: 1. After balancing the following equation for the combination of propane and oxygen, interpret the coefficients in terms of a) moles and b) liters C 3 H 8 + O 2 → CO 2 + H 2 O Solution : We can supply the following coefficients to obtain a balanced chemical equation. C 3 H 8 + 5 O 2 →3 CO 2 + 4 H 2 O

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11 TYPES OF STOICHIOMETRY

12 Mole- mole Problems A type of calculation that relates the moles of two substances participating in a balanced chemical equation. Consider the complete combustion of natural gas. Methane, CH 4, reacts with oxygen to give carbon dioxide and water. The equation is : CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O (g)

13 a. How many moles of oxygen react with 2.25 moles of CH 4 2.25 mol CH 4 X 2 mol O 2 = 4.50 mol O 2 1 mol CH 4 2. Calculate the moles of carbon dioxide produced from the reaction. 2.25 mol CH 4 X 1 mol CO 2 = 2.25 mol CO 2 1 mol CH 4

14 In this type of problem, we can convert from moles of moles of one substance to moles of another using a single unit. This type of conversion is sometimes referred to as a MOLE- MOLE PROBLEM.

15 MASS- MASS STOICHIOMETRY PROBLEMS

16 A mass-mass stoichiometry -an unknown mass of a substance is calculated from a given mass of a reactant or product in a chemical equation. Mass of Moles ofMoles of mass of Given→ Given→ Unknown → unknown

17 Consider the high temperature reduction of 14.4 g iron (II) oxide to elemental iron with aluminum metal. 3 FeO + 2 Al → 3 Fe + Al 2 O 3 14.4 g FeO X 1 mol FeO = 0.201 mol FeO 71.8 g FeO 0.201 mol FeO x 2 mol Al = 0.134 mol Al 3 mol FeO

18 0.134 mol Al x 27.0 g Al = 3.61 g Al 1 mol Al 14.4 g FeO X 1 mol FeO x 2 mol Al x 27.0 g Al = 3.61 g Al 71.8 g FeO 3 mol FeO 1 mol Al

19 Mass ofMoles ofMoles of gaseous volume Given → Given → Unknown → unknown Gaseous Moles of Moles of mass of Volume of → given→ Unknown → Unknown Given Vol MASS-VOLUME STOICHIOMETRY

20 The reaction of 0.165 g of Al metal reacts with dilute HCl forming aqueous AlCl 3 and Hydrogen gas. Find the volume of H2 at STP. 0.165 g Al x 1 mol Al x 3 mol H 2 x 22.4 L H 2 x 1000 ml H 2 = 205mL 27.0 g Al 2 mol Al 1 mol H 2 1 L H 2

21 VOLUME –VOLUME STOICHIOMETRY Sulfur dioxide gas is converted to sulfur trioxide using heat and platinum catalyst. The sulfur trioxide gas is then passed through water to produced sulfuric acid. The process is extremely important, since each year more sulfuric acid is used than any other single chemical. The balanced equation for the conversion of SO 2 to SO 3 is 2 SO 2 + O 2 → 2 SO 3

22 Calculate the volume of oxygen gas that react with 37.5 L of sulfur dioxide. This volume-volume problem requires only one step : 37.5 L SO 2 x 1 L O 2 = 18.8 L O 2 2 L SO 2


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