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Bellwork November 5 th

Bellwork November 5 th. Dig around in your brain and fill in the following… 1 mole = ___________________ atoms 1 mole = ___________________ liters 1 mole = ___________________ g H 2 O 3 mole = ________________ molecules. CHEMISTRY-1 CHAPTER 9. Stoichiometry.

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Bellwork November 5 th

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  1. BellworkNovember 5th • Dig around in your brain and fill in the following… • 1 mole = ___________________ atoms • 1 mole = ___________________ liters • 1 mole = ___________________ g H2O • 3 mole = ________________ molecules

  2. CHEMISTRY-1 CHAPTER 9 Stoichiometry

  3. Just What is a Chemical Ratio? • A mole ratio is like making a s’more. • The “chemical reaction for a s’more would be… • 3 chocolate squares + 1 marshmallow + 2 graham crackers  1 s’more • If you want to make 3 s’mores, how many marshmallows would you need? • 3 s’mores = 3 marshmallows • If you have 27 chocolate squares, how many smores can you make? • 3 chocolate squares = 1 s’more so 27 chocolate squares = 9 s’mores

  4. 9.1 2Mg + O2 2MgO Stoichiometry – study of calculations of quantities in chemical reactions using balanced chemical equations. 2 moles Mg + 1mole O2 2 moles MgO

  5. 2Mg + O2 2MgO • The mole ratios can be obtained from the coefficients in the balanced chemical equation. • What are the mole ratios in this problem? • Mole ratios can be used as conversion factors to predict the amount of any reactant or product involved in a reaction if the amount of another reactant and/or product is known.

  6. Mole Ratio! What’s that mean? Well, a stoichiometry problem gives you an amount of one chemical and asks you to solve for a different chemical. To get from one type of chemical to another, a MOLE RATIO must be found between the two chemicals. You get the MOLE RATIO from the BALANCED CHEMICAL EQUATION!

  7. The mole map R.P. of substance given R.P. of substance wanted Coefficients in a balanced equation! Atoms, formula units, molecules Atoms, formula units, molecules 6.02 x 1023 1mol 6.02 x 1023 1mol Mass of substance given Moles of substance given Moles of substance wanted Mass of substance wanted MM 1mol MM 1mol grams mol mol grams 22.4 L 1mol 22.4 L 1mol Always follow the road map… Do not make up new paths! Volume of substance given Volume of substance wanted L, mL L, mL

  8. A balanced chemical equation tells the quantity of reactants and products as well as what they are. 2 mol 1 mol 2 mol *the coefficients are* Mole Ratios

  9. The MOLE RATIO is your mechanism of transition between the chemical that is your starting given and the chemical you are solving for. The MOLE RATIO is the bridge between the two different chemicals moles moles (given) (want) Mole Ratio ? want given

  10. Mole - Mole Calculations – Mole ratios are used to calculate the number of moles of product from the given number of moles of reactant, or vice versa. 9.2 • Lets get back to those s’mores. If you have the following reaction… • 3 chocolate squares + 1 marshmallow + 2 graham crackers  1 s’more • You have 325 chocolate squares and you want to make the maximum number of s’mores possible. Show a calculation for this reaction. • 325 c.s. 1 s’more = 108.3333 or 108 s’mores • 3 c.s.

  11. EXAMPLE • Calculate the number of moles of Na2O that will be produced when 5.00 moles of Na completely react with oxygen gas. 4Na + O2 2Na2O 5.00 mol Na 2 mol Na2O = 2.50 mol Na2O 4 mol Na mol-mol ratio

  12. Mole-Mass Calculations: • We use mole ratios from balanced chemical equations and the molar conversions from Ch. 7 to calculate amounts (grams) of substances needed or produced in chemical reactions.

  13. EXAMPLE • How many grams of KClO3 must decompose to produce KCl and 1.45 moles O2? 2KClO3→ 2KCl + 3O2 2 mol KClO3 122.548 g KClO3 1.45 moles O2 = 3 mol O2 1 mol KClO3 118 g KClO3 mol-mol ratio MM

  14. Mass-Mass Calculations: • Same as mole-mass calculations, with an additional step of converting mass to moles.

  15. EXAMPLE • When 24.0 g of Na are mixed with Cl2, are 52.0 g of NaCl produced? Explain. 2 Na + Cl2→ NaCl 2 1 mol Na 2 mol NaCl 58.443 g NaCl 24.0 g Na = 22.990 g Na 2 mol Na 1 mol NaCl MM mol-mol ratio MM 61.0 g NaCl Are 52.0 g produced? Actually more than 52.0 g were produced.

  16. CaCO3, limestone, is heated to produce calcium oxide, CaO, and CO2. What mass of limestone is required to produce 156.0 g of CaO? CaCO3 (s)  CaO (s) + CO2 (g) 156.0 g CaO 1 mol CaO 1 mol CaCO3 100.086 g CaCO3 56.077 g CaO 1 mol CaO 1 mol CaCO3 MM mol-mol ratio MM 278.4 g CaCo3 =

  17. BellworkNovember 6th • Magnesium metal reacts with oxygen in a synthesis reaction. Write and balance the equation, then fill in the chart concerning the particles and mass.

  18. BellworkNovember 7th • A chemist combines 6.32g of C2H2 and 12.2g of oxygen. How many grams of carbon dioxide are produced?

  19. A quick reminder… • Let’s quickly remind ourselves of just how awesome a mole is! • 1 mole = ________________particles • 1 mole = ________________ liters • 1 moles = _______________ grams 6.02 x 1023 22.4 The mass from the periodic table!

  20. Combination Calculations: • At STP, how many grams of oxygen are needed to produce 19.8 liters SO3 according to the balanced equation below? 2SO2 (g) + O2 (g)  2SO3 (g) 19.8 L SO3 1 mol SO3 1 mol O2 31.998 g O2 22.4 L SO3 2 mol SO3 1 mol O2 Volume @ STP mol-mol ratio MM 14.1 g O2 =

  21. BellworkTuesday, November 10th • A student drops a 3.40g piece of zinc into hydrochloric acid. How many liters of hydrogen gas are produced if the reaction occurs at STP?

  22. BellworkTuesday, November 10th • How many grams of aluminum sulfate are produced if 23.33 g Al reacts with 74.44 g CuSO4?

  23. 9.3 Reactants in Excess and Limiting Reactants • Limiting Reactant - the reactant that is used up in a reaction. When you run out of a reactant, the reaction stops and no more product is formed. • Excess Reactant - the reactant that is not used up in a chemical reaction. • The limiting reactantdetermines the amount of product formed. Important

  24. Back to S’mores • The s’mores can teach us one more lesson; the lesson of limiting reagents. Lets look at our balanced equation one more time. • 3 chocolate squares + 1 marshmallow + 2 graham crackers  1 s’more • If your friend is throwing a slumber party and has 137 chocolate squares and 47 marshmallows, how many total smores can be made. • You will need to do the calculation with BOTH starting reactants. • 137 choc.squares 1 s’more = 45.6 s’mores • 3 choc.squares • 47 marshmallows 1 s’more = 47 s’mores • 1 marshmallow • So, in this case, what LIMITS our production of s’mores? • Chocolate squares! So chocolate squares are the limiting reagent!

  25. Steps to Determining the Limiting Reactant • Find out how much could be produced from each amount of reactant. • Which ever produces less is limiting the amount produced. It is therefore the Limiting Reactant.

  26. EXAMPLE Which amount produced is less? • If 6.70 mol of Na reacts with 3.20 mol Cl2, what is the limiting reagent? How many moles of NaCl will be produced? Na + Cl2→ NaCl 2 2 6.70 mol Na 2 mol NaCl = 6.70 mol NaCl 2 mol Na 3.20 mol Cl2 2 mol NaCl 6.40 mol NaCl = 1 mol Cl2 This is the amount produced, and determines which is the limiting reactant Cl2 is the limiting reactant

  27. To determine how much excess reactant is left after a reaction, subtract how much of the excess reactant reacted from how much excess reactant you started with.

  28. EXAMPLE • When 0.500 mole of aluminum reacts with 0.720 mole of iodine, to form aluminum iodide, AlI3(s), what mass of AlI3 is produced? How much of the excess reactant will remain? • I will separate this problem into two different steps. • Find the limiting reactant and amount produced (as in previous example) • Find amount in excess (left over)

  29. Al + I2→ AlI3 • When 0.500 mole of aluminum reacts with 0.720 mole of iodine, to form aluminum iodide, AlI3(s), what mass of AlI3 is produced? 2 3 2 0.500 mol Al 0.720 mol I2 2 mol AlI3 407.7 g AlI3 204 g AlI3 = 2 mol Al 1 mol AlI3 2 mol AlI3 407.7 g AlI3 = 196 g AlI3 3 mol l2 1 mol AlI3 amount produced I2 Limiting Reactant?

  30. First…take the Limiting Reactant and the amount with which you began… Al + I2→ AlI3 2 3 2 …then convert to find the amount that reacted with it… • How much of the excess reactant will remain? 0.720 mol I2 2 mol Al 0.480 mol Al = 3 mol l2 …then subtract the amount used from the amount given (starting amount) 0.500 mol Al - 0.480 mol Al = 0.020 mol Al given in the problem this is the excess (left over)

  31. BellworkWednesday, November 11th • 5.9 L of carbon dioxide is combined with 8.4 g MgO in a synthesis reaction to form magnesium carbonate. How many grams of magnesium carbonate is created in this reaction?

  32. Percent Yield • Theoretical Yield – The maximum amount of product that could be formed from given amounts of reactants. (What you should get) (You calculate this using dimensional analysis) • Actual Yield - amount of product actually obtained in the reaction (What you really get) (You either get this in a lab, or are given it in a problem)

  33. Percent Yielddescribes how much product was actually made in the lab versus the amount that theoretically could be made. Percent yield tells you how close you were to the 100% mark. • Reactions do not always work perfectly. Experimental error (spills, contamination) often means that the amount of product made in the lab does not match the ideal amount that could have been made.

  34. Amount you actually obtain in an experiment % yield = Actual Yield x 100 Theoretical Yield Amount you should obtain in an experiment (this we will calculate)

  35. EXAMPLE 12.5 g of copper are reacted with an excess of chlorine gas, then 25.4 g of copper(II) chloride are obtained. Calculate the theoretical yield and the percent yield. First write the balanced equation, then find the theoretical yield (amount you “should get”)

  36. First write the balanced equation, then find the theoretical yield (amount you “should get”) 12.5 g Cu Cu + Cl2→ CuCl2 1 mol Cu 1 mol CuCl2 134.452 g CuCl2 63.546gCu 1 mol Cu 1 mol CuCl2 MM mol-mol ratio MM = 26.4 g CuCl2 this is the theoretical yield

  37. Now do the % yield calculation % yield = Actual Yield x 100 Theoretical Yield given in problem 25.4 g CuCl2 % yield = X 100 = 96.2 % 26.4 g CuCl2 from calculation

  38. When I was a sophomore in college, we had a lab where we were supposed to isolate caffeine from tea leaves. Most of my caffeine was washed down the drain in a freak accident. Although I should have had 5.0 g of caffeine, I only ended up with 0.040 g of caffeine and a bad grade on the lab. What was my percent yield? 0.040 g X 100 = 0.80 % 5.0 g

  39. What is the theoretical yield if 5.50 grams of hydrogen react with nitrogen to form ammonia? H2 + N2 NH3 3 2 5.50 g H2 1 mole H2 17.031 g NH3 2 moles NH3 1 mol NH3 2.016 g H2 3 moles H2 = 31.2 grams NH3 = Theoretical Yield!!!! Only 20.4 grams of ammonia is actually produced in the lab. What is the percent yield? 20.4 g NH3 = 65.4 % yield X 100 31.2 g NH3

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