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STOICHIOMETRY – Chapter 11

STOICHIOMETRY – Chapter 11. What is stoichiometry?. Stoichiometry is the quantitative study of reactants and products in a chemical reaction. What You Should Expect. Given : Amount of reactants Question: how much of products can be formed. Example 2 A + 2B 3C

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STOICHIOMETRY – Chapter 11

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  1. STOICHIOMETRY – Chapter 11

  2. What is stoichiometry? • Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

  3. What You Should Expect • Given : Amount of reactants • Question:how much of products can be formed. • Example • 2 A + 2B 3C • Given 20.0 grams of A and sufficient B, how many grams of C can be produced?

  4. What do you need? You will need to use • molar ratios, • molar masses, • balancing and interpreting equations, and • conversions between grams and moles. Note: This type of problem is often called "mass-mass."

  5. Steps Involved in Solving Mass-Mass Stoichiometry Problems • Balance the chemical equation correctly • Using the molar mass of the given substance, convert the mass given to moles. • Construct a molar proportion (two molar ratios set equal to each other) • Using the molar mass of the unknown substance, convert the moles just calculated to mass.

  6. Mole Ratios A mole ratio converts moles of one compound in a balanced chemical equation into moles of another compound.

  7. Example Reaction between magnesium and oxygen to form magnesium oxide. ( fireworks) 2 Mg(s) + O2(g) 2 MgO(s) Mole Ratios: 2 : 1 : 2

  8. Practice Problems 1) N2 + 3 H2 ---> 2 NH3 Write the mole ratios for N2 to H2 and NH3 to H2. 2) A can of butane lighter fluid contains 1.20 moles of butane (C4H10). Calculate the number of moles of carbon dioxide given off when this butane is burned.

  9. Mole-Mole Problems Using the practice question 2) above: Equation of reaction 2C4H10 + 13O2 8CO2 + 10H2O Mole ratio C4H10 CO2 1 : 4 [ bases] 1.2 : X [ problem] By cross-multiplication, X = 4.8 mols of CO2 given off

  10. Mole-Mass Problems • Problem 1: 1.50 mol of KClO3 decomposes. How many grams of O2 will be produced? [k = 39, Cl = 35.5, O = 16] 2 KClO3 2 KCl + 3 O2

  11. Three steps…Get Your Correct Answer • Use mole ratio • Get the answer in moles and then • Convert to Mass. [Simple Arithmetic] Hello! If you are given a mass in the problem, you will need to convert this to moles first. Ok?

  12. Let’s go! 2 KClO3 2 KCl + 3 O2 2 : 3 1.50 : X X = 2.25mol Convert to mass 2.25 mol x 32.0 g/mol = 72.0 grams Cool!

  13. Try This: • We want to produce 2.75 mol of KCl. How many grams of KClO3 would be required? Soln KClO3 : KCl 2 : 2 X : 2.75 X = 2.75mol In mass: 2.75mol X 122.55 g/mol = 337 grams zooo zimple!

  14. Mass-MassProblems There are four steps involved in solving these problems: • Make sure you are working with a properly balanced equation. • Convert grams of the substance given in the problem to moles. • Construct two ratios - one from the problem and one from the equation and set them equal. Solve for "x," which is usually found in the ratio from the problem. • Convert moles of the substance just solved for into grams.

  15. Mass-VolumeProblems Just follow mass-mass problem to the penultimate level

  16. Like this: There are four steps involved in solving these problems: • Make sure you are working with a properly balanced equation. • Convert grams of the substance given in the problem to moles. • Construct two ratios - one from the problem and one from the equation and set them equal. Solve for "x," which is usually found in the ratio from the problem. • Convert moles of the substance just solved for into Volume.

  17. Conversion of mole to volume No of moles = Volume Molar volume Can you remember a similar equation?

  18. Molarvolume The molar volume is the volume occupied by one mole of ideal gas at STP. Its value is: 22.4dm3

  19. Practice Problems Calculate the volume of carbon dioxide formed at STP in ‘dm3' by the complete thermal decomposition of 3.125 g of pure calcium carbonate (Relative atomic mass of Ca=40, C=12, O=16) Solution: Convert the mass to mole: Molar mass of CaCO3 = 40 + 12 + (16 x 3) = 100gmol-1 Mole = mass/molar mass 3.125/100 = 0.03125mol

  20. Practice Problems As per the equation, Mole ratio 1 : 1 problem 0.03125mol X X = 0.03125mol of CO2 Convert mole to volume [slide 17] Volume = (0.03125 x 22.4)dm3 =0.7dm3

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