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Unit 9 part 2:

Unit 9 part 2:. Stoichiometry. 9.2.1 Stoichiometry. What is Stoichiometry? The study of the quantitative relationships that exist in chemical formulas and reactions. Stoicheion - element Metron - measure. Interpreting Balanced Chemical Equations.

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Unit 9 part 2:

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  1. Unit 9 part 2: Stoichiometry

  2. 9.2.1 Stoichiometry • What is Stoichiometry? • The study of the quantitative relationships that exist in chemical formulas and reactions. • Stoicheion- element • Metron- measure

  3. Interpreting Balanced Chemical Equations • To balance chemical equations we use coefficients- whole numbers in front of the formula. • These coefficient represent MOLE ratios for the substances in the rxn. • Ex.- N2H4 +2H2O2→ N2 + 4H2O means… 1mol N2H4 +2mol H2O2→ 1mol N2 + 4mol H2O

  4. Mole-Mole Problems • Knowing that the coefficients show MOLE RATIOS… we can now solve problems relating moles of one substance to moles of another. • Ex • N2H4 +2H2O2→ N2 + 4H2O How many mols of N2H4 are needed to react with 23 mols of H2O2 ?

  5. More mol- mol practice •   2S  +  3O2→2SO3 How many moles of S are needed to react with 2.5 mols of O2? • Cu + 4HNO3→Cu(NO3)2 + 2NO2 + 2H2O How many moles of Cu(NO3)2 are produced from 12.3 moles of HNO3?

  6. Verifying the Law of Conservation of Matter • Now we can PROVE balanced equations obey the law of conservation of matter. 2H2 + O2→ 2H2O We do this by calculating the mass of the reactant and comparing them to the mass of the products. H- 2molX 2g = 4g plus O- 1mol X 32g= 32g equals 36g H20- 2mol X 18g = 36g

  7. 9.2.2: Solving Stoichiometry Problems • There are 3 types of Stoichiometry problems: • 1. Mass-Mass problems • 2. Mass-Volume problems • 3. Volume-Volume problems In general, every problem will be solved in 3 steps: Quantity of given →mols of given →mols of unknown → quantity of unknown

  8. Mass-Mass Problems • In these problems, you will be given the mass of one substance and asked to solve for the mass of another substance. • Your mantra is: • Mass to moles, moles to moles, moles to mass. • You will need a balanced chem. equation, the molar mass of the knownand the unknownsubstance to solve.

  9. Mass-mass practice • 2H2 + O2→ 2H2O • How many grams of water will be produced from 23.5 g of oxygen? • Mass to moles, moles to moles, moles to mass • Convert g to moles of O2, then mol O2 to mol H2O, and finally, mols of H2O to g of H2O.

  10. More practice • Zn + H2SO4→ZnSO4 + H2 • How many g of H2SO4 will react with 9.5 g of Zn? • HCl + NaOH → NaCl + H2O • How many g of NaCl will be produced from 51.2 g of NaOH? • 2Mg + O2→ 2MgO • How many g of Mg are needed to produce 12.3g MgO?

  11. Mass-Volume Problems In these problems you will be given the MASS of one substance and asked to find the VOLUME of a gas. You will be going from grams to liters (g → L). Your Mantra is: Mass to moles, moles to moles, moles to volume. To convert from mols to volume you will use the molar volume (22.4L/ mol).

  12. Mass-Volume practice • 2H2 + O2→ 2H2O • How many liters of hydrogen will be required to produce 23.5 g of water? • Mass to moles, moles to moles, moles to volume. • Convert g to moles of H2O, then mol H2O to mol H2, and finally, mols of H2 to L of H2. • Remember, use the molar volume: 22.4 L/mol

  13. More practice • Zn + H2SO4→ZnSO4 + H2 • How many L of H2 will be produced from 9.5 g of Zn? • Sn + 2HF → SnF2 + H2 • How many g of HF are needed to react with 30.0L of H2? • NH4NO3→ N2O + 2H2O • How many L of N2O will be produced from 52.6 g NH4NO3? • 2Mg + O2→ 2MgO • How many g of MgO are needed to produce 12.3L O2?

  14. Volume-Volume Problems In these problems you will be given the VOLUME of one gas and asked to find the VOLUME of a different gas. Your Mantra is: VOLUME to moles, moles to moles, moles to volume. To convert from mols to volume you will use the molar volume (22.4L/ mol).

  15. Volume - Volume practice • 2H2 + O2→ 2H2O • How many liters of hydrogen will be required to react with 23.5 L of O2? • volume to moles, moles to moles, moles to volume. • Convert L to moles of O2, then mol O2 to mol H2, and finally, mols of H2 to L of H2. • Remember, use the molar volume: 22.4 L/mol

  16. More practice • Sn + 2HF → SnF2 + H2 • How many L of HF will be required to form 3.00L of H2? • N2 + 3H2→2NH3 • How many L of N2 will be required to react with 5.2 L H2?

  17. 9.3.3 Limiting Reactants and Percent Yields • The amount of product made depends on how much reactants are available. Sometimes, one of the reactants limits the number of products made… • How many bikes can be made from 11 bike frames but only 7 tires? • Did the # of bike frames or the # of tires limit how many bikes were made? • The reactant that limits the amount of product formed is called the limiting reactant. • The reactant that is not completely used up is called the excess reactant.

  18. Identifying Limiting Reactants • There are 3 general steps to identifying the limiting reactant. • 1. Begin with a balanced chemical equation. • 2. Calculate the amount of product formed by EACH of the reactants. (pick only one product) • 3. The reactant that produces the least amount of product is the limiting reactant.

  19. Limiting Reactant Problems • Cu + 2AgNO3→ Cu(NO3)2 + 2Ag • Which is the limiting reactant if you have 6.0g Cu and 12.5g AgNO3 ? (hint: solve for Ag) • You will perform 2 mass-mass problems. • 1. mass of Cu to mass of Ag • 2. mass of AgNO3 to mass of Ag.

  20. Percent Yield The amount we calculate and what we actually make as the product are not necessarily the same amount. Expected yield – amount of product that should be produced based on calculation. Actual yield – the amount ACTUALLY obtained from the reaction (in lab).

  21. We need to know how much of the expected was made during the reaction. …did we only make 5% or 55%? • Percent Yield – what % of the predicted amount was actually made. % yield = (actual yield ÷ expected yield) X 100%

  22. Percent Yield practice problems • You burn Mg in O2 and produce 6.5g MgO. You expected to make 8.2g of MgO. What is your percent yield? • Determine the %yield for 3.74g Na and excess O2 if 5.34g of Na2O2 is recovered? We don’t know the expected yield so you will have to calculate it first, then solve %yield.

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