1 / 25

Stoichiometry

Stoichiometry. Chemistry 6.0. The Mathematics of Chemical Reactions: STOICHIOMETRY. I. Balanced Chemical Equations A. Provide qualitative and quantitative information B. Supports the Law of Conservation of Matter 2H 2 + O 2  2H 2 O

gasha
Download Presentation

Stoichiometry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Stoichiometry Chemistry 6.0

  2. The Mathematics of Chemical Reactions: STOICHIOMETRY I. Balanced Chemical Equations A. Provide qualitative and quantitative information B. Supports the Law of Conservation of Matter 2H2 + O2 2H2O The above equation is interpreted in terms of particles as follows: 1. 2 moleculesof H2 react with 1 moleculeof O2 to produce 2 moleculesof water. The ratio of H2 to O2 is 2:1. or 2 molesof H2 react with 1 moleof O2 to produce 2 moles of water. The ratio of H2 to O2 is 2:1.

  3. 2. It is more convenient to interpret the coefficients as number of moles, because we measure amounts of substances by massing. C. Stoichiometry 1. The study of the quantitative relationships that exist in a formula or a chemical reaction. 2. Importance a. Provides for the safe, economical and reproducible manufacture of chemicals. b. Provides for the safe administration of pharmaceuticals. D. Proof of the conservation of matter in a balanced equation 1. Convert all reactants and products to their mass equivalents. 2. Sum up the mass of reactants and compare the sum of the mass of products.

  4. II. Stoichiometry Problems A. Steps to Solve Problems • Write a balanced equation. • Identify the given ( ) and the unknown or required substance (?). • Convert mass of given into moles. • Use the mole (molar) ratio to convert from given to required substance. • If needed, convert moles of required into mass of required substance.

  5. B. Examples 1. How many moles of oxygen are required to react with 16 moles of hydrogen in the production of water? H2 + O2 H2O  ? 2 2 16 moles H2 1 mole O2 2 moles H2 8.0 moles O2 = Mole ratio links 1 substance to another in a reaction. Required in problem solving.

  6. Antimony reacts with water to produce antimony(III) oxide • and hydrogen. How many moles of hydrogen are produced • from 7.5 moles of antimony? Sb + H2O  Sb2O3 + H2 2 3 3 11 mol H2

  7. What mass of aluminum oxide can be prepared by the reaction of 67.5 g of aluminum in a synthesis reaction? Al + O2 Al2O3 4 3 2 128 mol Al2O3

  8. Sodium bicarbonate, a.k.a. baking soda, can be used to extinguish a fire. When heated, it decomposes to give carbon dioxide gas which smothers the fire. It also produces sodium carbonate and water. If a sample contains 4.0 g of sodium bicarbonate, what mass of carbon dioxide is produced? • 2 NaHCO3 Na2CO3 + H2O + CO2

  9. III. Percent Yield A. Expected Yield: the amount of product that should be produced (theoretical) B. Actual Yield: the amount of product that is actually produced (experimental) C. Percent Yield: percent of expected yield that was obtained % Yield = (actual yield/expected yield) x 100

  10. D. Steps to Solving Percent Yield Problems • Write a balanced equation • Identify the given () which is the mass of reactant, and identify the actual yield. • Solve for the expected mass of product using the given mass of reactant. • Calculate the % yield. % Yield = actual yield x 100 expected yield

  11. E. Examples 1. A reaction between 2.80 g aluminum nitrate and excess sodium hydroxide produced 0.966 g of aluminum hydroxide in this double replacement reaction. Calculate the % yield. Al(NO3)3 + 3NaOH  Al(OH)3 + 3NaNO3 1.03 g Al(OH)3

  12. 2. Determine the % yield for the reaction between 3.74 g of sodium and excess oxygen if 4.24 g of sodium oxide is recovered in the direct combination reaction. 4Na + O2 2Na2O

  13. IV. Limiting Reactants • Definition: the reactant that determines, or limits, the amount of product(s) formed in a chemical reaction • Problem Solving Tips • The limiting reactant is not necessarily the reactant present in the smallest amount • When you are given the amounts of 2 or more reactants, you should suspect that you are dealing with a limiting reactant problem.

  14. C. Steps • Write a balanced equation • Calculate the number of moles of each reactant • Compare the mole ratios of the reactants as available ratio (from the given masses) and the required ratio (from the coefficients) • Identify the limiting reactant, and use it to calculate the mass of product formed.

  15. D. Examples 1. What mass of CO2 could be formed by the combustion of 16.0 g CH4 with 48.0 g O2? CH4 + 2O2 CO2 + 2H2O 33.0 g CO2 Oxygen is the limiting reactant

  16. 2. What is the maximum mass of nickel(II) hydroxide that could be prepared by mixing 25.9 g nickel(II) chloride with 10.0 g sodium hydroxide? NiCl2 + 2NaOH  Ni(OH)2 + 2NaCl 11.6 g Ni(OH)2 Sodium Hydroxide is the LR

  17. V. Sequential Reactions • Definition: A chemical process in which several reactions are required to convert starting materials into product(s). • Main Concept: The amount of desired product from each reaction is taken as the starting material for the next reaction. Consideration is given to steps that produce less than 100%.

  18. C. Examples 1. Given the following sequence of equations, calculate the mass of Ni(CO)4 produced from 75.0 g of carbon. Assume 100% yields. C + H2O  CO + H2 Ni + 4CO  Ni(CO)4

  19. 2. Hydrogen, obtained by the electrical decomposition of water, is combined with chlorine to produce 84.2 g of hydrogen chloride. Calculate the mass of water decomposed. Assume 100% yield. 2H2O  2H2 + O2 H2 + Cl2 2HCl

  20. VI. Solution Stoichiometry A. Many reactants are introduced to a reaction chamber as a solution. B. The most common solution concentration is molarity. molarity = mol/liter C. Examples • Excess lead(II) carbonate reacts with 27.5 mL of 3.00M nitric acid. Calculate the mass of lead(II) nitrate formed PbCO3 + 2HNO3 Pb(NO3)2 + H2CO3

  21. 2. Calculate the volume, in mL, of a 0.324 molar solution of sulfuric acid required to react completely with 2.792 g of sodium carbonate according to the equation below. H2SO4 + Na2CO3 Na2SO4 + CO2 + H2O

  22. Problems C3H8 + 5O2 → 3CO2 + 4H2O ΔH = -2.22x103 kJ • How much heat is released when 22.0g of propane is burned? • How much carbon dioxide is produced, in grams, when 2,500 kJ of energy is released? -1.11x103 kJ (released) 150 g CO2

  23. ΔH from ΔHf The standard enthalpy change of a reaction is equal to the sum of the standard molar enthalpies of formation of the products multiplied by its coefficient, n, in the balanced equation, minus the corresponding sum of standard molar enthalpies of formation of reactants. Hrxn = ∑ nHf,products - ∑ nHf,reactants ***By definition, the standard enthalpy of formation of the most stable form of any element is zero because there is no formation reaction needed when the element is already in its standard state.***

  24. ΔH from ΔHf Problem: Using the Heats of Formation Table, calculate the H for the following reaction: SF6(g) + 3H2O(l)  6HF(g) + SO3(g) Write the thermochemical equation for this reaction: SF6 + 3H2O + 45.1kJ  6HF + SO3 45.1 kJ

  25. ΔH from ΔHf Problem: Using the Heats of Formation Table, calculate the standard heat of combustion for propane. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) Write the thermochemical equation for this reaction: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) + 2043.9kJ -2043.9 kJ

More Related