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- 1. Chapter 12 - Stoichiometry I ‘m back! Objective: To learn how to use a completechemical equation to calculate quantities of a substance
- 2. Chemical Equations Remember that a balanced chemical equation provides the same kind of quantitative information that a recipe does. From the specific amounts of the ingredients, a specific amount of cookies can be made. Due to this, proportional relationships b/w the ingredients and the products can be written. What would happen if you add more or less of 1 ingredient?
- 3. Proportional Relationships2 1/4 c. flour 3/4 c. brown sugar1 tsp. baking soda 1 tsp vanilla extract1 tsp. salt 2 eggs1 c. butter 2 c. chocolate chips3/4 c. sugar Makes 5 dozen cookies. 1) 1 tsp. baking soda = 2 c. chocolate chips 2) 1 c butter= 2 eggs 3) 5 dz cookies = 1tsp. Salt 4) 5 dz cookies = ¾ c. sugar ETC.
- 4. Proportional Relationships2 1/4 c. flour 3/4 c. brown sugar1 tsp. baking soda 1 tsp vanilla extract1 tsp. salt 2 eggs1 c. butter 2 c. chocolate chips3/4 c. sugar Makes 5 dozen cookies.I have 5 eggs. How many cookies can I make? Ratio of eggs to cookies 5 eggs 5 doz. = 12.5 dozen cookies 2 eggs
- 5. Stoichiometryx Stoichiometry - The study of quantities of materials consumed & produced in chemical reactions. • Uses a balanced chemical equation • Similar to bookkeeping. • Based on MOLE RATIOS
- 6. Interpreting Balanced Equations 2H2 + O2 → 2H2O 2 molecules of hydrogen and 1 molecule ofoxygen form 2 molecules of water. 2 dozen molecules hydrogen and 1 dozenmolecules of oxygen form 2 dozen moleculesof water. 2 moles of hydrogen and 1 mole of oxygenform 2 moles of water.
- 7. Mole Ratio tells the ratio between 2 substancesin a balanced chemical equation. • indicated by coefficients in a balanced equation • Ex. 2 Mg + O2 → 2 MgO 2 moles Mg 1 mole O2 or 2 moles Mg 1 mole O2 2 moles MgO 2 mole Mg 2 mole Mg or 2 moles MgO 2 moles MgO 1 mole O2 or 1 mole O2 2 moles MgO
- 8. LAW of CONSERVATION of MASSThe MOLE relationships can also apply to massrelationships. We can check this converting from gramsto moles. 2H2 + O2 → 2H2OReactants: 2.02 g H2 2 moles H2 = 4.04 g H2 1 mole H21 mole O2 32.00 g O2 = 32.00 g O2 36.04 g H2 + O2 1 mole O2Product: 18.02 g H2O = 36.04 g H2O2 moles H2O 1 mole H O 2
- 9. Steps to Calculate Stoichiometric Problems1. Correctly balance the equation.2. Convert the given amount into moles.3. Set up mole ratios.4. Use mole ratios to calculate moles of desired chemical.5. Convert moles back into final unit.
- 10. Types of CalculationsRemember the mole ratio is the ratio betweenany two substances in a reaction and will beused in all calculations1. mole to moleEx. 1 How many moles of O2 are produced when3.34 moles of Al2O3 decompose? 2 Al2O3 → 4Al + 3O23.34 mol Al2O3 3 mol O2 = 5.01 mol O2 2 mol Al2O3
- 11. *Practice problem 1: C2H2 + O2 → CO2 + H2O (unbalanced)If 3.84 moles of C2H2 are burned, how manymoles of O2 are needed?
- 12. types of problems (continued)2. moles to grams given moles mole ratio molar mass Ex. 2 If 1.50 moles of Fe are used , how many grams of copper II sulfate would form? 2Fe + 3CuSO4 → Fe2(SO4)3 + 3Cu1.50 mol of Fe 3 mol CuSO4 159.62 g CuSO4 2 mol Fe 1 mol CuSO4 = 359 g CuSO4
- 13. *Practice problem 2: C2H2 + O2 → CO2 + H2OIf 0.750 moles of O2 are used, how many grams ofC2H2 are burned?3. grams to molesgiven mass molar mass mole ratioEx. 3 If 146 g of NaCl are produced in the followingreaction, how many moles of Cl2 were consumed?2Na + Cl2 2NaCl146 g NaCl 1 mol NaCl 1 mol Cl2 = 1.25 mol 58.44 g NaCl 2 mol NaCl
- 14. *Practice problem 3: Calculate the no. of molesof CH4 used to produce 1.8 g of H2O. CH4 + 2O2 → CO2 + 2H2O types of problems (continued)4. grams to gramsgiven mass molar mass mole ratio molar massEx. 4 If 10.8 g of Fe are added to a solution ofCuSO4, how many grams of solid copper wouldform? 2Fe + 3CuSO4 → Fe2(SO4)3 + 3Cu10.8 g Fe 1 mol Fe 3 mol Cu 63.55 g Cu = 18.4g 55.85 g Fe 2 mol Fe 1 mol Cu
- 15. *Practice problem 4: Zinc reacts withiodine to form zinc iodide (ZnI2). Write thebalanced equation for this reaction.Calculate the grams of ZnI2 that would beproduced from a 125.0 g sample of Zn.

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