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Stoichiometry

Learn about stoichiometry, the branch of chemistry that deals with mass relationships in compounds and the balance of chemical equations. Explore topics such as mole ratios, conversions, mass-mass calculations, limiting reagents, and percentage yield.

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Stoichiometry

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  1. The branch of chemistry that deals with the mass relationships of elements in compounds and the mass relationships between reactants and products in a chemical equation Stoichiometry is like following a recipe Focuses on mass relationships Stoichiometry * The key is a balanced equation and reading the equation in terms of…Coefficients!

  2. Na + Cl  ______ Type of reaction? Skeleton equation Balanced equation: *Check for diatomic elements! *Check charges on ionic compounds! *Count the atoms and balance! How is this read in terms of moles? What are all the possible mole ratios? “RECIPE” for moles to moles, moles to mass, mass to moles and mass to mass Let’s try this

  3. Conversions of Quantities in Moles • CO2 (g) + 2LiOH(s) → Li2CO3(s) + H2O (l) How many moles of lithium hydroxide are required to react with 20 moles of carbon dioxide, the average amount exhaled by a person each day? List the givens and unknowns! 2 moles/1mole = x/20moles Answer = 40 moles

  4. Practice Problems 1 and 2 on page 306 Remember! Start with the skeleton equation and then balance the equation before calculating an answer! Do Now The pain! The pain!

  5. Conversions of Amounts in Moles to Mass • Plants use energy from the sun to produce glucose and oxygen from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when 3.00 moles of water react with carbon dioxide? 1) Write the skeleton equation. • Write the balanced equation. • List the given and unknown. • Calculate the mole ratio. • Multiply the mole ratio by the molar mass of the unknown. Answer: 90.1 grams

  6. Question cont’d • What mass of carbon dioxide, in grams, is needed to react with 3.00 moles of water? More Problems Do practice problems 1 and 2 on page 308

  7. Conversions of Mass to Amounts in Moles • The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3 (g) + 02 (g) → NO (g) + H2O (g) (unbalanced) • This reaction is run using 824 g NH3 and excess oxygen. 1. How many moles of NO are formed? 2. How many moles of H2O are formed?

  8. Solve Question 1 • Balance the equation. • List the given and unknown. • Find the molar mass of NH3. • Find the mole ratio of NO to NH3. • Convert the given mass to moles of the given. • Find the answer using the mole ratio. 4NH3 (g) + 502 (g) → 4NO (g) + 6H2O (g) Given: mass of NH3 = 824 grams Unknown: moles of NO produced 1 mole of NH3 = 17.04 grams 4:4 ratio is 1:1 ratio 824/17.04 = 48.4 mol NH3 1:1 ratio, so 48.4 mol NO

  9. Solve Question 2 • Balance the equation. • List the given and unknown. • Find the molar mass of NH3. • Find the mole ratio of H2O to NH3. • Convert the given mass to moles of the given. • Find the answer using the mole ratio. 4NH3 (g) + 502 (g) → 4NO (g) + 6H2O (g) Given: mass of NH3 = 824 grams Unknown: moles of H2O produced 1 mole of NH3 = 17.04 grams 6:4 ratio is 3:2 ratio 824/17.04 = 48.4 mol NH3 [6/4 = x/48.4 or 3/2 = x/48.4] Answer = 72.6 moles

  10. Mass-Mass Calculations • Tin(II) fluoride, SnF2 , is used in some toothpastes. It is made by the reaction of tin with hydrogen fluoride according to the following equation. Sn(s) + 2HF(g) → SnF2(s) + H2(g) How many grams of SnF2 are produced from the reaction of 30.00 g HF with Sn?

  11. The equation is balanced so list the given and unknown. • Calculate the molar masses of HF and SnF2. • Calculate the mole ratio of HF to SnF2. • Calculate moles of HF based on given grams. • Use the mole ratio and plug in the calculated moles of HF. • Multiply the number of moles of SnF2 by the molar mass of SnF2. Molar mass of HF = 20.01 g Molar mass of SnF2 = 156.71 g Mole ratio is 2:1 HF: 30 g/20.01 g = 1.499 mol 2/1 = 1.499/x, so x = .7495 .7495 x 156.71 = 117.45 g, or 117.5 g Answer: 117.5 g

  12. Limiting Reagent • Once a reactant is used up, there can be no more products produced • The substance that is used up first is called the limiting reagent = the reactant that limits the amount of the other reactant that can be combined and thus the product that is produced • Excess reagent = the substance not used up completely • Look at Figure 5 on page 312 in the text; then look at Sample Problems F and G (Pages 313-314) • Take Notes!

  13. Percentage Yield • Theoretical yield = what you calculate/expect under perfect conditions • Actual Yield = what you get in the lab • Percentage Yield = (AY/TY) x 100% • Look at Sample Problem H (p. 317)

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